ChemistryQ301–350 of 806 questions
Page 7 of 12 · English
301
ChemistryMediumMCQMHT CET · 2026
What type of hybridization is found in $[Ni(Cl)_4]^{2-}$?
A
$sp^3$
B
$sp^3d^2$
C
$d^2sp^3$
D
$dsp^2$
Solution
(A) $1$. The central metal ion is $Ni^{2+}$. The atomic number of $Ni$ is $28$, so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$2$. $Cl^-$ is a weak field ligand, so it does not cause pairing of electrons in the $3d$ orbitals.
$3$. The $3d$ orbitals remain as $3d^8$ with two unpaired electrons.
$4$. To accommodate four $Cl^-$ ligands, the metal ion uses one $4s$ and three $4p$ orbitals.
$5$. Thus, the hybridization is $sp^3$, resulting in a tetrahedral geometry.
$2$. $Cl^-$ is a weak field ligand, so it does not cause pairing of electrons in the $3d$ orbitals.
$3$. The $3d$ orbitals remain as $3d^8$ with two unpaired electrons.
$4$. To accommodate four $Cl^-$ ligands, the metal ion uses one $4s$ and three $4p$ orbitals.
$5$. Thus, the hybridization is $sp^3$, resulting in a tetrahedral geometry.
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302
ChemistryMediumMCQMHT CET · 2026
Identify the ionization isomer of $[Cr(H_2O)_4 Cl(NO_2)] Cl$ from the following.
A
$[Cr(H_2O)_4 (NO_2)] Cl_2$
B
$[Cr(H_2O)_4 Cl_2](NO_2)$
C
$[Cr(H_2O)_4 Cl(ONO)] Cl$
D
$[Cr(H_2O)_4 Cl(NO_2)] H_2O$
Solution
(B) Step $1$: Ionization isomers are compounds that produce different ions in solution. This occurs when a counter ion in the coordination compound exchanges its position with a ligand present in the coordination sphere.
Step $2$: The given complex is $[Cr(H_2O)_4 Cl(NO_2)] Cl$. Here, $Cl^-$ is the counter ion.
Step $3$: To form an ionization isomer, the $Cl^-$ ion outside the coordination sphere must exchange with a ligand inside the sphere (e.g., $NO_2^-$).
Step $4$: Swapping $Cl^-$ with $NO_2^-$ gives the complex $[Cr(H_2O)_4 Cl_2](NO_2)$.
Step $5$: Thus, option $B$ is the correct ionization isomer.
Step $2$: The given complex is $[Cr(H_2O)_4 Cl(NO_2)] Cl$. Here, $Cl^-$ is the counter ion.
Step $3$: To form an ionization isomer, the $Cl^-$ ion outside the coordination sphere must exchange with a ligand inside the sphere (e.g., $NO_2^-$).
Step $4$: Swapping $Cl^-$ with $NO_2^-$ gives the complex $[Cr(H_2O)_4 Cl_2](NO_2)$.
Step $5$: Thus, option $B$ is the correct ionization isomer.
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303
ChemistryMediumMCQMHT CET · 2026
Which of the following complexes is an example of the type $MA_4BC$ diastereoisomers?
A
$[Pt(NH_3)_4 ClBr]^{2+}$
B
$[Pt(NH_3)_4 Cl_2]$
C
$[Co(NH_3)_4 Cl_2]^{+}$
D
$[Pt(NH_3)_2 Cl_2]$
Solution
(A) Step $1$: Identify the general formula $MA_4BC$. Here, $M$ is the central metal atom, $A$ is a monodentate ligand present in $4$ units, and $B$ and $C$ are two different monodentate ligands.
Step $2$: Analyze the options. Option $A$ is $[Pt(NH_3)_4 ClBr]^{2+}$. Here, $M = Pt$, $A = NH_3$ ($4$ units), $B = Cl$, and $C = Br$. This matches the $MA_4BC$ type.
Step $3$: Diastereoisomers (geometric isomers) for $MA_4BC$ type complexes exist in two forms: $cis$ (where $B$ and $C$ are adjacent) and $trans$ (where $B$ and $C$ are opposite).
Step $4$: Therefore, $[Pt(NH_3)_4 ClBr]^{2+}$ can exhibit geometric isomerism.
Step $2$: Analyze the options. Option $A$ is $[Pt(NH_3)_4 ClBr]^{2+}$. Here, $M = Pt$, $A = NH_3$ ($4$ units), $B = Cl$, and $C = Br$. This matches the $MA_4BC$ type.
Step $3$: Diastereoisomers (geometric isomers) for $MA_4BC$ type complexes exist in two forms: $cis$ (where $B$ and $C$ are adjacent) and $trans$ (where $B$ and $C$ are opposite).
Step $4$: Therefore, $[Pt(NH_3)_4 ClBr]^{2+}$ can exhibit geometric isomerism.
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304
ChemistryDifficultMCQMHT CET · 2026
Identify the formula of $\text{Bis(ethylenediamine)dithiocyanatoplatinum(IV) ion}$ from the following.
A
$[Pt(en)_2(NCS)_2]^{2+}$
B
$[Pt(en)_2(SCN)_2]^{4+}$
C
$[Pt(en)_2(SCN)_2]^{2+}$
D
$[Pt(en)_3(SCN)_2]^{4+}$
Solution
(C) $1$. The central metal ion is $\text{Platinum}$ in oxidation state $+IV$, denoted as $Pt^{4+}$.
$2$. $\text{Bis(ethylenediamine)}$ indicates two $en$ ligands, which are neutral ($0$ charge).
$3$. $\text{Dithiocyanato}$ indicates two $SCN^-$ ligands, each with a $-1$ charge.
$4$. The total charge on the complex ion is calculated as: $\text{Charge} = \text{Oxidation state of } Pt + (2 \times \text{charge of } en) + (2 \times \text{charge of } SCN^-)$.
$5$. $\text{Charge} = (+4) + (2 \times 0) + (2 \times -1) = +4 - 2 = +2$.
$6$. Thus, the formula is $[Pt(en)_2(SCN)_2]^{2+}$.
$2$. $\text{Bis(ethylenediamine)}$ indicates two $en$ ligands, which are neutral ($0$ charge).
$3$. $\text{Dithiocyanato}$ indicates two $SCN^-$ ligands, each with a $-1$ charge.
$4$. The total charge on the complex ion is calculated as: $\text{Charge} = \text{Oxidation state of } Pt + (2 \times \text{charge of } en) + (2 \times \text{charge of } SCN^-)$.
$5$. $\text{Charge} = (+4) + (2 \times 0) + (2 \times -1) = +4 - 2 = +2$.
$6$. Thus, the formula is $[Pt(en)_2(SCN)_2]^{2+}$.
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305
ChemistryMediumMCQMHT CET · 2026
In a complex compound, the central metal ion acts as a:
A
Arrhenius acid
B
Arrhenius base
C
Lewis acid
D
Lewis base
Solution
(C) $1$. In a coordination complex, the central metal ion has vacant orbitals.
$2$. Ligands donate a pair of electrons to these vacant orbitals to form coordinate covalent bonds.
$3$. According to the Lewis theory, an electron pair acceptor is defined as a Lewis acid.
$4$. Since the central metal ion accepts electron pairs from ligands, it acts as a Lewis acid.
$2$. Ligands donate a pair of electrons to these vacant orbitals to form coordinate covalent bonds.
$3$. According to the Lewis theory, an electron pair acceptor is defined as a Lewis acid.
$4$. Since the central metal ion accepts electron pairs from ligands, it acts as a Lewis acid.
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306
ChemistryMediumMCQMHT CET · 2026
Which of the following sets of ligands does $NOT$ contain an ambident ligand?
A
Oxalate ion, $NO_2^{-}, NCS^{-}$
B
$C_2O_4^{2-}$, Ethylene diamine, $H_2O$
C
$NO_2^{-}, C_2O_4^{2-}, EDTA^{4-}$
D
$EDTA^{4-}, NCS^{-}$, Oxalate ion
Solution
(B) $1$. An ambident ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$2$. Common examples of ambident ligands include $NO_2^{-}$ (can bond via $N$ or $O$) and $NCS^{-}$ (can bond via $N$ or $S$).
$3$. In option $(A)$, $NO_2^{-}$ and $NCS^{-}$ are ambident ligands.
$4$. In option $(B)$, $C_2O_4^{2-}$ (oxalate), ethylene diamine $(en)$, and $H_2O$ are all polydentate or monodentate ligands but none of them are ambident.
$5$. In option $(C)$, $NO_2^{-}$ is an ambident ligand.
$6$. In option $(D)$, $NCS^{-}$ is an ambident ligand.
$7$. Therefore, the set that does not contain any ambident ligand is $(B)$.
$2$. Common examples of ambident ligands include $NO_2^{-}$ (can bond via $N$ or $O$) and $NCS^{-}$ (can bond via $N$ or $S$).
$3$. In option $(A)$, $NO_2^{-}$ and $NCS^{-}$ are ambident ligands.
$4$. In option $(B)$, $C_2O_4^{2-}$ (oxalate), ethylene diamine $(en)$, and $H_2O$ are all polydentate or monodentate ligands but none of them are ambident.
$5$. In option $(C)$, $NO_2^{-}$ is an ambident ligand.
$6$. In option $(D)$, $NCS^{-}$ is an ambident ligand.
$7$. Therefore, the set that does not contain any ambident ligand is $(B)$.
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307
ChemistryMediumMCQMHT CET · 2026
Which of the following is an ambident ligand?
A
Ethylene diamine
B
$Cl^{-}$
C
$EDTA^{4-}$
D
$SCN^{-}$
Solution
(D) $1$. An ambident ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$2$. $SCN^{-}$ (thiocyanate ion) can coordinate through the sulfur atom $(S)$ to form a thiocyanato complex or through the nitrogen atom $(N)$ to form an isothiocyanato complex.
$3$. Ethylene diamine is a bidentate ligand, $Cl^{-}$ is a monodentate ligand, and $EDTA^{4-}$ is a hexadentate ligand.
$4$. Therefore, $SCN^{-}$ is an ambident ligand.
$2$. $SCN^{-}$ (thiocyanate ion) can coordinate through the sulfur atom $(S)$ to form a thiocyanato complex or through the nitrogen atom $(N)$ to form an isothiocyanato complex.
$3$. Ethylene diamine is a bidentate ligand, $Cl^{-}$ is a monodentate ligand, and $EDTA^{4-}$ is a hexadentate ligand.
$4$. Therefore, $SCN^{-}$ is an ambident ligand.
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308
ChemistryEasyMCQMHT CET · 2026
Among the given ligands, the negative ligand is
A
$NH_3$
B
$CH_3NH_2$
C
$N_2H_5^+$
D
$NO_3^-$
Solution
(D) $1$. $A$ ligand is an ion or molecule that binds to a central metal atom to form a coordination complex.
$2$. $NH_3$ (Ammonia) is a neutral ligand.
$3$. $CH_3NH_2$ (Methylamine) is a neutral ligand.
$4$. $N_2H_5^+$ (Hydrazinium) is a positively charged ligand.
$5$. $NO_3^-$ (Nitrate) is a negatively charged ligand (anionic ligand).
$6$. Therefore, the correct option is $D$.
$2$. $NH_3$ (Ammonia) is a neutral ligand.
$3$. $CH_3NH_2$ (Methylamine) is a neutral ligand.
$4$. $N_2H_5^+$ (Hydrazinium) is a positively charged ligand.
$5$. $NO_3^-$ (Nitrate) is a negatively charged ligand (anionic ligand).
$6$. Therefore, the correct option is $D$.
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309
ChemistryDifficultMCQMHT CET · 2026
The sum of the coordination number and the oxidation number of $M$ in $[M(en)_2C_2O_4]Cl$ is
A
$9$
B
$8$
C
$7$
D
$6$
Solution
(A) $1$. The coordination number of $M$ is calculated by considering the denticity of the ligands. $en$ (ethylenediamine) is a bidentate ligand and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
$2$. Coordination number $= (2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
$3$. Let the oxidation number of $M$ be $x$. The charge on $en$ is $0$, the charge on $C_2O_4^{2-}$ is $-2$, and the charge on $Cl^-$ is $-1$.
$4$. The total charge of the complex $[M(en)_2C_2O_4]Cl$ is $0$. Thus, $x + 2(0) + 1(-2) - 1 = 0$.
$5$. $x - 2 - 1 = 0 \implies x = +3$.
$6$. The sum of the coordination number and the oxidation number $= 6 + 3 = 9$.
$2$. Coordination number $= (2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
$3$. Let the oxidation number of $M$ be $x$. The charge on $en$ is $0$, the charge on $C_2O_4^{2-}$ is $-2$, and the charge on $Cl^-$ is $-1$.
$4$. The total charge of the complex $[M(en)_2C_2O_4]Cl$ is $0$. Thus, $x + 2(0) + 1(-2) - 1 = 0$.
$5$. $x - 2 - 1 = 0 \implies x = +3$.
$6$. The sum of the coordination number and the oxidation number $= 6 + 3 = 9$.
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310
ChemistryMediumMCQMHT CET · 2026
Identify the oxidation state of the cobalt ion in the complex $[Co(NH_3)_5Br]SO_4$.
A
$+2$
B
$+3$
C
$+1$
D
$+4$
Solution
(B) $1$. The complex $[Co(NH_3)_5Br]SO_4$ dissociates into $[Co(NH_3)_5Br]^{2+}$ and $SO_4^{2-}$.
$2$. Let the oxidation state of $Co$ be $x$.
$3$. The oxidation state of $NH_3$ is $0$ and $Br$ is $-1$.
$4$. The charge on the complex ion $[Co(NH_3)_5Br]^{2+}$ is $+2$.
$5$. Setting up the equation: $x + 5(0) + (-1) = +2$.
$6$. Solving for $x$: $x - 1 = +2$, which gives $x = +3$.
$2$. Let the oxidation state of $Co$ be $x$.
$3$. The oxidation state of $NH_3$ is $0$ and $Br$ is $-1$.
$4$. The charge on the complex ion $[Co(NH_3)_5Br]^{2+}$ is $+2$.
$5$. Setting up the equation: $x + 5(0) + (-1) = +2$.
$6$. Solving for $x$: $x - 1 = +2$, which gives $x = +3$.
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311
ChemistryEasyMCQMHT CET · 2026
The number of ligands which are directly bonded to the metal is known as
A
co-ordination number
B
oxidation number
C
co-ordination sphere
D
valency
Solution
(A) The co-ordination number of a metal atom in a complex is defined as the number of ligand donor atoms to which the metal is directly bonded. Therefore, the correct option is $A$.
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312
ChemistryMediumMCQMHT CET · 2026
Identify the total number of complexes from the following list that contain only monodentate ligands:
$(a)$ $[Ni(CN)_4]^{2-}$ (Tetracyanonickelate $(II)$ ion)
$(b)$ $K_3[AlF_6]$ (Potassium hexafluoroaluminate $(III)$)
$(c)$ $[Cu(NH_3)_4]^{2+}$ (Tetraamminecopper$(II)$ ion)
$(d)$ $K_3[Al(C_2O_4)_3]$ (Potassium trioxalatoaluminate $(III)$)
$(a)$ $[Ni(CN)_4]^{2-}$ (Tetracyanonickelate $(II)$ ion)
$(b)$ $K_3[AlF_6]$ (Potassium hexafluoroaluminate $(III)$)
$(c)$ $[Cu(NH_3)_4]^{2+}$ (Tetraamminecopper$(II)$ ion)
$(d)$ $K_3[Al(C_2O_4)_3]$ (Potassium trioxalatoaluminate $(III)$)
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(B) Step $1$: Analyze the ligands in each complex.
$(a)$ $[Ni(CN)_4]^{2-}$: The ligand is $CN^-$, which is a monodentate ligand.
$(b)$ $[AlF_6]^{3-}$: The ligand is $F^-$, which is a monodentate ligand.
$(c)$ $[Cu(NH_3)_4]^{2+}$: The ligand is $NH_3$, which is a monodentate ligand.
$(d)$ $[Al(C_2O_4)_3]^{3-}$: The ligand is $C_2O_4^{2-}$ (oxalate), which is a bidentate ligand.
Step $2$: Count the complexes containing only monodentate ligands. Complexes $(a)$, $(b)$, and $(c)$ contain only monodentate ligands. Complex $(d)$ contains a bidentate ligand.
Step $3$: The total number of such complexes is $3$.
$(a)$ $[Ni(CN)_4]^{2-}$: The ligand is $CN^-$, which is a monodentate ligand.
$(b)$ $[AlF_6]^{3-}$: The ligand is $F^-$, which is a monodentate ligand.
$(c)$ $[Cu(NH_3)_4]^{2+}$: The ligand is $NH_3$, which is a monodentate ligand.
$(d)$ $[Al(C_2O_4)_3]^{3-}$: The ligand is $C_2O_4^{2-}$ (oxalate), which is a bidentate ligand.
Step $2$: Count the complexes containing only monodentate ligands. Complexes $(a)$, $(b)$, and $(c)$ contain only monodentate ligands. Complex $(d)$ contains a bidentate ligand.
Step $3$: The total number of such complexes is $3$.
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313
ChemistryDifficultMCQMHT CET · 2026
What is the Effective Atomic Number $(EAN)$ of $Co$ in $[Co(NH_3)_6]^{3+}$?
A
$24$
B
$27$
C
$36$
D
$30$
Solution
(C) $1$. The atomic number $(Z)$ of $Co$ is $27$.
$2$. The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is calculated as: $x + 6(0) = +3$, so $x = +3$.
$3$. The number of electrons in $Co^{3+}$ ion is $27 - 3 = 24$.
$4$. Each $NH_3$ ligand donates $2$ electrons. Since there are $6$ ligands, total electrons donated = $6 \times 2 = 12$.
$5$. $EAN = (Z - \text{oxidation state}) + (2 \times \text{coordination number}) = 24 + 12 = 36$.
$2$. The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is calculated as: $x + 6(0) = +3$, so $x = +3$.
$3$. The number of electrons in $Co^{3+}$ ion is $27 - 3 = 24$.
$4$. Each $NH_3$ ligand donates $2$ electrons. Since there are $6$ ligands, total electrons donated = $6 \times 2 = 12$.
$5$. $EAN = (Z - \text{oxidation state}) + (2 \times \text{coordination number}) = 24 + 12 = 36$.
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314
ChemistryDifficultMCQMHT CET · 2026
What is the Effective Atomic Number $(EAN)$ of the metal ion in $[Cu(NH_3)_4]^{2+}$?
A
$35$
B
$27$
C
$29$
D
$36$
Solution
(A) Step $1$: Identify the atomic number $(Z)$ of $Cu$ $(Z = 29)$.
Step $2$: Determine the oxidation state of $Cu$ in $[Cu(NH_3)_4]^{2+}$. Let it be $x$. $x + 4(0) = +2$, so $x = +2$.
Step $3$: Calculate the number of electrons in the metal ion: $Z - \text{oxidation state} = 29 - 2 = 27$.
Step $4$: Calculate the number of electrons donated by ligands: $4 \times 2 = 8$.
Step $5$: $EAN = (Z - \text{oxidation state}) + \text{electrons donated by ligands} = 27 + 8 = 35$.
Step $2$: Determine the oxidation state of $Cu$ in $[Cu(NH_3)_4]^{2+}$. Let it be $x$. $x + 4(0) = +2$, so $x = +2$.
Step $3$: Calculate the number of electrons in the metal ion: $Z - \text{oxidation state} = 29 - 2 = 27$.
Step $4$: Calculate the number of electrons donated by ligands: $4 \times 2 = 8$.
Step $5$: $EAN = (Z - \text{oxidation state}) + \text{electrons donated by ligands} = 27 + 8 = 35$.
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315
ChemistryDifficultMCQMHT CET · 2026
Find the Effective Atomic Number $(EAN)$ of $Fe$ in $Fe(CO)_5$.
A
$35$
B
$36$
C
$26$
D
$30$
Solution
(B) The formula for $EAN$ is: $EAN = Z - O.S. + 2 \times C.N.$
Where:
$Z$ (Atomic number of $Fe$) = $26$
$O.S.$ (Oxidation state of $Fe$ in $Fe(CO)_5$) = $0$ (since $CO$ is a neutral ligand)
$C.N.$ (Coordination number of $Fe$) = $5$
$EAN = 26 - 0 + 2 \times 5$
$EAN = 26 + 10 = 36$
Where:
$Z$ (Atomic number of $Fe$) = $26$
$O.S.$ (Oxidation state of $Fe$ in $Fe(CO)_5$) = $0$ (since $CO$ is a neutral ligand)
$C.N.$ (Coordination number of $Fe$) = $5$
$EAN = 26 - 0 + 2 \times 5$
$EAN = 26 + 10 = 36$
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316
ChemistryDifficultMCQMHT CET · 2026
What is the Effective Atomic Number $(EAN)$ for the metal ion in $[Fe(CN)_6]^{3-}$?
A
$37$
B
$36$
C
$35$
D
$34$
Solution
(C) Step $1$: Identify the atomic number $(Z)$ of the central metal ion $Fe$. $Z = 26$.
Step $2$: Determine the oxidation state of $Fe$ in $[Fe(CN)_6]^{3-}$. Let it be $x$. $x + 6(-1) = -3$, so $x = +3$.
Step $3$: Calculate the number of electrons in the $Fe^{3+}$ ion. $26 - 3 = 23$.
Step $4$: Calculate the number of electrons donated by $6$ ligands $(CN^-)$. Each $CN^-$ donates $2$ electrons. $6 \times 2 = 12$.
Step $5$: Calculate $EAN = (Z - \text{oxidation state}) + (2 \times \text{coordination number}) = 23 + 12 = 35$.
Step $2$: Determine the oxidation state of $Fe$ in $[Fe(CN)_6]^{3-}$. Let it be $x$. $x + 6(-1) = -3$, so $x = +3$.
Step $3$: Calculate the number of electrons in the $Fe^{3+}$ ion. $26 - 3 = 23$.
Step $4$: Calculate the number of electrons donated by $6$ ligands $(CN^-)$. Each $CN^-$ donates $2$ electrons. $6 \times 2 = 12$.
Step $5$: Calculate $EAN = (Z - \text{oxidation state}) + (2 \times \text{coordination number}) = 23 + 12 = 35$.
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317
ChemistryMediumMCQMHT CET · 2026
What is the oxidation state of cobalt in $[Co(NH_3)_6]^{3+}$?
A
$+5$
B
$+4$
C
$+3$
D
$+2$
Solution
(C) Let the oxidation state of cobalt $(Co)$ be $x$.
The oxidation state of the neutral ligand ammonia $(NH_3)$ is $0$.
The overall charge on the complex ion $[Co(NH_3)_6]^{3+}$ is $+3$.
Setting up the equation: $x + 6(0) = +3$.
Solving for $x$: $x = +3$.
Therefore, the oxidation state of cobalt is $+3$.
The oxidation state of the neutral ligand ammonia $(NH_3)$ is $0$.
The overall charge on the complex ion $[Co(NH_3)_6]^{3+}$ is $+3$.
Setting up the equation: $x + 6(0) = +3$.
Solving for $x$: $x = +3$.
Therefore, the oxidation state of cobalt is $+3$.
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318
ChemistryMediumMCQMHT CET · 2026
Which of the following coordination complexes contains a neutral ligand?
A
$[Fe(CO)_5]$ (Pentacarbonyliron$(0)$)
B
$[Fe(CN)_6]^{4-}$ (Hexacyanoferrate$(II)$ ion)
C
$[Co(C_2O_4)_3]^{3-}$ (Trioxalatocobaltate$(III)$ ion)
D
$Na_3[Co(NO_2)_6]$ (Sodium hexanitrocobaltate$(III)$)
Solution
(A) $1$. Identify the ligands in each complex:
- In $[Fe(CO)_5]$, the ligand is $CO$ (carbonyl), which is a neutral ligand.
- In $[Fe(CN)_6]^{4-}$, the ligand is $CN^-$ (cyano), which is an anionic ligand.
- In $[Co(C_2O_4)_3]^{3-}$, the ligand is $C_2O_4^{2-}$ (oxalato), which is an anionic ligand.
- In $[Co(NO_2)_6]^{3-}$, the ligand is $NO_2^-$ (nitro), which is an anionic ligand.
$2$. Since $CO$ is a neutral molecule, $[Fe(CO)_5]$ contains a neutral ligand.
$3$. Therefore, the correct option is $A$.
- In $[Fe(CO)_5]$, the ligand is $CO$ (carbonyl), which is a neutral ligand.
- In $[Fe(CN)_6]^{4-}$, the ligand is $CN^-$ (cyano), which is an anionic ligand.
- In $[Co(C_2O_4)_3]^{3-}$, the ligand is $C_2O_4^{2-}$ (oxalato), which is an anionic ligand.
- In $[Co(NO_2)_6]^{3-}$, the ligand is $NO_2^-$ (nitro), which is an anionic ligand.
$2$. Since $CO$ is a neutral molecule, $[Fe(CO)_5]$ contains a neutral ligand.
$3$. Therefore, the correct option is $A$.
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319
ChemistryMediumMCQMHT CET · 2026
What is the coordination number of the central metal ion in the $[Co(en)_3]^{3+}$ complex?
A
$2$
B
$4$
C
$6$
D
$8$
Solution
(C) $1$. The complex is $[Co(en)_3]^{3+}$, where $Co^{3+}$ is the central metal ion.
$2$. The ligand $en$ stands for ethylenediamine $(NH_2CH_2CH_2NH_2)$, which is a bidentate ligand.
$3$. $A$ bidentate ligand donates two lone pairs of electrons to the central metal ion.
$4$. The coordination number is calculated as: $\text{Number of ligands} \times \text{denticity} = 3 \times 2 = 6$.
$5$. Therefore, the coordination number of the central metal ion is $6$.
$2$. The ligand $en$ stands for ethylenediamine $(NH_2CH_2CH_2NH_2)$, which is a bidentate ligand.
$3$. $A$ bidentate ligand donates two lone pairs of electrons to the central metal ion.
$4$. The coordination number is calculated as: $\text{Number of ligands} \times \text{denticity} = 3 \times 2 = 6$.
$5$. Therefore, the coordination number of the central metal ion is $6$.
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320
ChemistryMediumMCQMHT CET · 2026
Which of the following coordination complexes is heteroleptic?
A
$[Fe(CO)_5]$
B
$[Co(NH_3)_3(NO_2)_3]$
C
$[Cu(NH_3)_4]^{2+}$
D
$[Ni(CN)_4]^{2-}$
Solution
(B) $1$. $A$ heteroleptic complex is a coordination complex in which the central metal atom or ion is bonded to more than one type of donor group (ligand).
$2$. In $[Fe(CO)_5]$, all ligands are $CO$.
$3$. In $[Co(NH_3)_3(NO_2)_3]$, the central metal $Co^{3+}$ is bonded to two different types of ligands: $NH_3$ and $NO_2^-$.
$4$. In $[Cu(NH_3)_4]^{2+}$, all ligands are $NH_3$.
$5$. In $[Ni(CN)_4]^{2-}$, all ligands are $CN^-$.
$6$. Therefore, $[Co(NH_3)_3(NO_2)_3]$ is the heteroleptic complex.
$2$. In $[Fe(CO)_5]$, all ligands are $CO$.
$3$. In $[Co(NH_3)_3(NO_2)_3]$, the central metal $Co^{3+}$ is bonded to two different types of ligands: $NH_3$ and $NO_2^-$.
$4$. In $[Cu(NH_3)_4]^{2+}$, all ligands are $NH_3$.
$5$. In $[Ni(CN)_4]^{2-}$, all ligands are $CN^-$.
$6$. Therefore, $[Co(NH_3)_3(NO_2)_3]$ is the heteroleptic complex.
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321
ChemistryMediumMCQMHT CET · 2026
Which of the following complexes is neutral?
A
$K_3[Fe(CN)_5NO]$
B
$[Zn(NH_3)_4]^{2+}$
C
$[Co(NH_3)_5Cl]^{2+}$
D
$[Ni(CO)_4]$
Solution
(D) complex is neutral if the net charge on the coordination entity is zero.
$(1)$ $K_3[Fe(CN)_5NO]$: The coordination entity is $[Fe(CN)_5NO]^{3-}$, which is charged.
$(2)$ $[Zn(NH_3)_4]^{2+}$: This is a cationic complex with a charge of $+2$.
$(3)$ $[Co(NH_3)_5Cl]^{2+}$: This is a cationic complex with a charge of $+2$.
$(4)$ $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$ and $CO$ is a neutral ligand. The net charge is $0 + 4(0) = 0$. Thus, it is a neutral complex.
$(1)$ $K_3[Fe(CN)_5NO]$: The coordination entity is $[Fe(CN)_5NO]^{3-}$, which is charged.
$(2)$ $[Zn(NH_3)_4]^{2+}$: This is a cationic complex with a charge of $+2$.
$(3)$ $[Co(NH_3)_5Cl]^{2+}$: This is a cationic complex with a charge of $+2$.
$(4)$ $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$ and $CO$ is a neutral ligand. The net charge is $0 + 4(0) = 0$. Thus, it is a neutral complex.
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322
ChemistryEasyMCQMHT CET · 2026
Which of the following is used as a solvent for the preparation of a $Grignard$ reagent?
A
Alcohol
B
Ether
C
Phenol
D
Ketone
Solution
(B) Step $1$: $Grignard$ reagents $(RMgX)$ are highly reactive organometallic compounds.
Step $2$: They react violently with protic solvents like water, alcohols, or acids to form alkanes.
Step $3$: Therefore, an anhydrous, aprotic solvent is required for their preparation.
Step $4$: Dry $Ether$ (diethyl ether) is the most commonly used solvent because it is inert towards the reagent and helps in stabilizing the $Grignard$ reagent through the formation of a coordinate bond between the oxygen atom of the ether and the magnesium atom.
Step $2$: They react violently with protic solvents like water, alcohols, or acids to form alkanes.
Step $3$: Therefore, an anhydrous, aprotic solvent is required for their preparation.
Step $4$: Dry $Ether$ (diethyl ether) is the most commonly used solvent because it is inert towards the reagent and helps in stabilizing the $Grignard$ reagent through the formation of a coordinate bond between the oxygen atom of the ether and the magnesium atom.
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323
ChemistryDifficultMCQMHT CET · 2026
Identify the product '$B$' in the following reaction sequence:
$CH_3CH_2CH(OH)CH_3 \xrightarrow{PBr_3} A \xrightarrow{Mg, \text{dry ether}} B$
$CH_3CH_2CH(OH)CH_3 \xrightarrow{PBr_3} A \xrightarrow{Mg, \text{dry ether}} B$
A
B
C
D
Solution
(C) Step $1$: Reaction of $butan-2-ol$ with $PBr_3$ replaces the $-OH$ group with $-Br$ to form $2-bromobutane$ (Product '$A$').
$CH_3CH_2CH(OH)CH_3 + PBr_3 \rightarrow CH_3CH_2CH(Br)CH_3 + H_3PO_3$
Step $2$: Reaction of $2-bromobutane$ with $Mg$ in the presence of dry ether forms the Grignard reagent, $sec-butylmagnesium$ $bromide$ (Product '$B$').
$CH_3CH_2CH(Br)CH_3 + Mg \xrightarrow{\text{dry ether}} CH_3CH_2CH(MgBr)CH_3$
The structure corresponding to $sec-butylmagnesium$ $bromide$ is given in option $C$.
$CH_3CH_2CH(OH)CH_3 + PBr_3 \rightarrow CH_3CH_2CH(Br)CH_3 + H_3PO_3$
Step $2$: Reaction of $2-bromobutane$ with $Mg$ in the presence of dry ether forms the Grignard reagent, $sec-butylmagnesium$ $bromide$ (Product '$B$').
$CH_3CH_2CH(Br)CH_3 + Mg \xrightarrow{\text{dry ether}} CH_3CH_2CH(MgBr)CH_3$
The structure corresponding to $sec-butylmagnesium$ $bromide$ is given in option $C$.
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324
ChemistryEasyMCQMHT CET · 2026
What is the importance of Sandmeyer's reaction?
A
To obtain $R-I$ from $R-Cl$
B
To obtain $Ar-X$ by replacing diazo group of $Ar-N_2^+X^-$ using $CuCl/HCl$
C
To obtain $R-F$ from $R-Cl$ using $AgF$
D
To obtain $R-CH_2-R$ from $R-CO-R$ using $Zn-Hg/HCl$
Solution
(B) $1$. Sandmeyer's reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts.
$2$. In this reaction, the diazonium group $(-N_2^+X^-)$ of an aryl diazonium salt is replaced by a halogen atom ($-Cl$, $-Br$, or $-CN$) using copper$(I)$ salts like $CuCl/HCl$ or $CuBr/HBr$.
$3$. Therefore, the correct option is the one describing the replacement of the diazo group to form $Ar-X$.
$2$. In this reaction, the diazonium group $(-N_2^+X^-)$ of an aryl diazonium salt is replaced by a halogen atom ($-Cl$, $-Br$, or $-CN$) using copper$(I)$ salts like $CuCl/HCl$ or $CuBr/HBr$.
$3$. Therefore, the correct option is the one describing the replacement of the diazo group to form $Ar-X$.
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325
ChemistryEasyMCQMHT CET · 2026
Identify the reagent used for the replacement of the $-N_2^+Cl^-$ group from benzene diazonium chloride by an iodine atom.
A
$HI$
B
$KOI$
C
$KI$
D
$PI_3$
Solution
(C) The reaction of benzene diazonium chloride with potassium iodide $(KI)$ is a standard method to introduce an iodine atom into the benzene ring.
Step $1$: The diazonium salt reacts with $KI$ in an aqueous solution.
Step $2$: The $-N_2^+Cl^-$ group is replaced by an iodine atom to form iodobenzene and nitrogen gas $(N_2)$ is evolved.
Reaction: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2 + KCl$.
Step $1$: The diazonium salt reacts with $KI$ in an aqueous solution.
Step $2$: The $-N_2^+Cl^-$ group is replaced by an iodine atom to form iodobenzene and nitrogen gas $(N_2)$ is evolved.
Reaction: $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2 + KCl$.
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326
ChemistryEasyMCQMHT CET · 2026
The reaction of an aryl halide with sodium metal in the presence of dry ether to form a biphenyl is known as:
A
Swartz reaction
B
Wurtz reaction
C
Fittig reaction
D
Wurtz-Fittig reaction
Solution
(C) Step $1$: The reaction involves two molecules of an aryl halide reacting with $2 \text{ mol}$ of sodium metal in the presence of dry ether.
Step $2$: The general equation is $2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$.
Step $3$: This specific coupling reaction of aryl halides is known as the Fittig reaction.
Step $4$: Therefore, the correct option is $C$.
Step $2$: The general equation is $2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX$.
Step $3$: This specific coupling reaction of aryl halides is known as the Fittig reaction.
Step $4$: Therefore, the correct option is $C$.
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327
ChemistryMediumMCQMHT CET · 2026
Which of the following $n$ mole compound is obtained when $2n$ moles of $C_6H_5Cl$ are treated with $2n$ moles of sodium atom in the presence of dry ether?
A
Biphenyl
B
Toluene
C
Phenol
D
Benzene
Solution
(A) The reaction described is the Wurtz-Fittig reaction.
In this reaction, an aryl halide reacts with an alkyl halide or another aryl halide in the presence of sodium metal and dry ether to form a coupled product.
The chemical equation for the reaction of $C_6H_5Cl$ with $Na$ is:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$.
Given $2n$ moles of $C_6H_5Cl$ and $2n$ moles of $Na$, the stoichiometry indicates that $n$ moles of $C_6H_5-C_6H_5$ (Biphenyl) will be formed.
In this reaction, an aryl halide reacts with an alkyl halide or another aryl halide in the presence of sodium metal and dry ether to form a coupled product.
The chemical equation for the reaction of $C_6H_5Cl$ with $Na$ is:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$.
Given $2n$ moles of $C_6H_5Cl$ and $2n$ moles of $Na$, the stoichiometry indicates that $n$ moles of $C_6H_5-C_6H_5$ (Biphenyl) will be formed.
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328
ChemistryMediumMCQMHT CET · 2026
The major product '$B$' in the below mentioned reaction is:
$CH_3-CH(Br)-CH_3 \xrightarrow[\Delta]{\text{Alc. KOH}} A \xrightarrow[\text{Peroxide absent}]{\text{HBr}} B$
$CH_3-CH(Br)-CH_3 \xrightarrow[\Delta]{\text{Alc. KOH}} A \xrightarrow[\text{Peroxide absent}]{\text{HBr}} B$
A
Bromoethane
B
$1-$Bromopropane
C
$2-$Bromopropane
D
Carbon tetrabromide
Solution
(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ leads to the formation of propene $(A)$ via an elimination reaction.
$CH_3-CH(Br)-CH_3 + \text{Alc. KOH} \xrightarrow{\Delta} CH_3-CH=CH_2 (A) + KBr + H_2O$
Step $2$: The addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule, where the negative part of the addendum $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide absent}} CH_3-CH(Br)-CH_3 (B)$
Thus, the major product '$B$' is $2$-bromopropane.
$CH_3-CH(Br)-CH_3 + \text{Alc. KOH} \xrightarrow{\Delta} CH_3-CH=CH_2 (A) + KBr + H_2O$
Step $2$: The addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule, where the negative part of the addendum $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide absent}} CH_3-CH(Br)-CH_3 (B)$
Thus, the major product '$B$' is $2$-bromopropane.
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329
ChemistryMediumMCQMHT CET · 2026
$A$ galvanic cell consists of a copper electrode and a standard hydrogen electrode. If $E^\circ (Cu^{2+}_{(aq)} | Cu_{(s)}) = +0.34 \text{ V}$, identify the reaction taking place at the positive electrode during the working of the cell.
A
$Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2e^-$
B
$Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$
C
$H_2(g) \longrightarrow 2H^+_{(aq)} + 2e^-$
D
$H^+_{(aq)} + 2e^- \longrightarrow H_2(g)$
Solution
(B) $1$. In a galvanic cell, the electrode with the higher reduction potential acts as the cathode (positive electrode).
$2$. The standard reduction potential of the standard hydrogen electrode $(SHE)$ is $E^\circ (H^+ | H_2) = 0.00 \text{ V}$.
$3$. Since $E^\circ (Cu^{2+} | Cu) = +0.34 \text{ V} > 0.00 \text{ V}$, the copper electrode acts as the cathode.
$4$. Reduction occurs at the cathode: $Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$.
$2$. The standard reduction potential of the standard hydrogen electrode $(SHE)$ is $E^\circ (H^+ | H_2) = 0.00 \text{ V}$.
$3$. Since $E^\circ (Cu^{2+} | Cu) = +0.34 \text{ V} > 0.00 \text{ V}$, the copper electrode acts as the cathode.
$4$. Reduction occurs at the cathode: $Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$.
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330
ChemistryMediumMCQMHT CET · 2026
During the electrolysis of aqueous $NaCl$ (brine), the product formed at the cathode is?
A
$Cl_{2(g)}$
B
$O_{2(g)}$
C
$Na_{(s)}$
D
$H_{2(g)}$
Solution
(D) $1$. Aqueous $NaCl$ contains $Na^+$, $Cl^-$, $H^+$, and $OH^-$ ions.
$2$. At the cathode, the reduction reaction with the higher standard reduction potential occurs.
$3$. The reduction potential of $H^+ + e^- \rightarrow \frac{1}{2} H_{2(g)}$ $(E^0 = 0.00 \text{ V})$ is higher than that of $Na^+ + e^- \rightarrow Na_{(s)}$ $(E^0 = -2.71 \text{ V})$.
$4$. Therefore, $H^+$ ions are reduced to $H_{2(g)}$ gas at the cathode.
$2$. At the cathode, the reduction reaction with the higher standard reduction potential occurs.
$3$. The reduction potential of $H^+ + e^- \rightarrow \frac{1}{2} H_{2(g)}$ $(E^0 = 0.00 \text{ V})$ is higher than that of $Na^+ + e^- \rightarrow Na_{(s)}$ $(E^0 = -2.71 \text{ V})$.
$4$. Therefore, $H^+$ ions are reduced to $H_{2(g)}$ gas at the cathode.
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331
ChemistryDifficultMCQMHT CET · 2026
What is the number of moles of electrons passed when a current of $2 \text{ A}$ flows through a solution of an electrolyte for $10 \text{ minutes}$?
A
$1.243 \times 10^{-2}$
B
$1.784 \times 10^{-2}$
C
$2.022 \times 10^{-2}$
D
$8.041 \times 10^{-2}$
Solution
(A) Step $1$: Calculate the total charge $(Q)$ passed using the formula $Q = I \times t$.
Given: $I = 2 \text{ A}$, $t = 10 \text{ minutes} = 10 \times 60 \text{ s} = 600 \text{ s}$.
$Q = 2 \text{ A} \times 600 \text{ s} = 1200 \text{ C}$.
Step $2$: Calculate the number of moles of electrons $(n)$ using the relation $n = \frac{Q}{F}$, where $F$ is Faraday's constant $(96500 \text{ C/mol})$.
$n = \frac{1200 \text{ C}}{96500 \text{ C/mol}} \approx 0.012435 \text{ mol}$.
Step $3$: Express in scientific notation: $n \approx 1.243 \times 10^{-2} \text{ mol}$.
Given: $I = 2 \text{ A}$, $t = 10 \text{ minutes} = 10 \times 60 \text{ s} = 600 \text{ s}$.
$Q = 2 \text{ A} \times 600 \text{ s} = 1200 \text{ C}$.
Step $2$: Calculate the number of moles of electrons $(n)$ using the relation $n = \frac{Q}{F}$, where $F$ is Faraday's constant $(96500 \text{ C/mol})$.
$n = \frac{1200 \text{ C}}{96500 \text{ C/mol}} \approx 0.012435 \text{ mol}$.
Step $3$: Express in scientific notation: $n \approx 1.243 \times 10^{-2} \text{ mol}$.
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332
ChemistryDifficultMCQMHT CET · 2026
What is the quantity of electricity in Faradays required to produce $8 \text{ g}$ of $Mg$ (molar mass = $24 \text{ g mol}^{-1}$) from $MgCl_2$ solution (in $\text{ F}$)?
A
$0.222$
B
$0.444$
C
$0.666$
D
$0.888$
Solution
(C) The chemical reaction for the reduction of $Mg^{2+}$ to $Mg$ is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.
From the stoichiometry, $1 \text{ mole}$ of $Mg$ requires $2 \text{ moles}$ of electrons, which is equal to $2 \text{ F}$ of electricity.
Molar mass of $Mg = 24 \text{ g mol}^{-1}$.
Number of moles of $Mg$ produced = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{8 \text{ g}}{24 \text{ g mol}^{-1}} = \frac{1}{3} \text{ mol}$.
Since $1 \text{ mole}$ of $Mg$ requires $2 \text{ F}$, then $\frac{1}{3} \text{ mole}$ of $Mg$ requires $\frac{1}{3} \times 2 \text{ F} = 0.666 \text{ F}$.
Therefore, the correct option is $C$.
From the stoichiometry, $1 \text{ mole}$ of $Mg$ requires $2 \text{ moles}$ of electrons, which is equal to $2 \text{ F}$ of electricity.
Molar mass of $Mg = 24 \text{ g mol}^{-1}$.
Number of moles of $Mg$ produced = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{8 \text{ g}}{24 \text{ g mol}^{-1}} = \frac{1}{3} \text{ mol}$.
Since $1 \text{ mole}$ of $Mg$ requires $2 \text{ F}$, then $\frac{1}{3} \text{ mole}$ of $Mg$ requires $\frac{1}{3} \times 2 \text{ F} = 0.666 \text{ F}$.
Therefore, the correct option is $C$.
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333
ChemistryDifficultMCQMHT CET · 2026
Determine the electrode potential of $Sn^{2+} (0.01 \text{ M}) | Sn(s)$ at $25^\circ \text{C}$ if $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$. (in $\text{ V}$)
A
$0.186$
B
$-0.186$
C
$-0.195$
D
$0.195$
Solution
(C) The half-cell reaction is: $Sn^{2+} (aq) + 2e^- \rightarrow Sn(s)$.
Using the Nernst equation: $E_{Sn^{2+}/Sn} = E^\circ_{Sn^{2+}/Sn} - \frac{0.0591}{n} \log \frac{1}{[Sn^{2+}]}$.
Here, $n = 2$, $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$, and $[Sn^{2+}] = 0.01 \text{ M} = 10^{-2} \text{ M}$.
$E_{Sn^{2+}/Sn} = -0.136 - \frac{0.0591}{2} \log \frac{1}{10^{-2}}$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.02955 \times \log(10^2)$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.02955 \times 2$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.0591 = -0.1951 \text{ V} \approx -0.195 \text{ V}$.
Using the Nernst equation: $E_{Sn^{2+}/Sn} = E^\circ_{Sn^{2+}/Sn} - \frac{0.0591}{n} \log \frac{1}{[Sn^{2+}]}$.
Here, $n = 2$, $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$, and $[Sn^{2+}] = 0.01 \text{ M} = 10^{-2} \text{ M}$.
$E_{Sn^{2+}/Sn} = -0.136 - \frac{0.0591}{2} \log \frac{1}{10^{-2}}$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.02955 \times \log(10^2)$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.02955 \times 2$.
$E_{Sn^{2+}/Sn} = -0.136 - 0.0591 = -0.1951 \text{ V} \approx -0.195 \text{ V}$.
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334
ChemistryDifficultMCQMHT CET · 2026
The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$
A
$18.02 \text{ kJ mol}^{-1}$
B
$20.1 \text{ kJ mol}^{-1}$
C
$19.02 \text{ kJ mol}^{-1}$
D
$21.5 \text{ kJ mol}^{-1}$
Solution
(C) Given: $k_1 = 1.5 \times 10^7 \text{ s}^{-1}$, $T_1 = 300 \text{ K}$, $k_2 = 3.0 \times 10^7 \text{ s}^{-1}$, $T_2 = 330 \text{ K}$.
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
$\log \left( \frac{3.0 \times 10^7}{1.5 \times 10^7} \right) = \frac{E_a}{19.15} \left( \frac{330 - 300}{300 \times 330} \right)$.
$\log(2) = \frac{E_a}{19.15} \left( \frac{30}{99000} \right)$.
$0.3010 = \frac{E_a}{19.15} \times \frac{1}{3300}$.
$E_a = 0.3010 \times 19.15 \times 3300 \text{ J mol}^{-1}$.
$E_a = 19023.225 \text{ J mol}^{-1} \approx 19.02 \text{ kJ mol}^{-1}$.
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
$\log \left( \frac{3.0 \times 10^7}{1.5 \times 10^7} \right) = \frac{E_a}{19.15} \left( \frac{330 - 300}{300 \times 330} \right)$.
$\log(2) = \frac{E_a}{19.15} \left( \frac{30}{99000} \right)$.
$0.3010 = \frac{E_a}{19.15} \times \frac{1}{3300}$.
$E_a = 0.3010 \times 19.15 \times 3300 \text{ J mol}^{-1}$.
$E_a = 19023.225 \text{ J mol}^{-1} \approx 19.02 \text{ kJ mol}^{-1}$.
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335
ChemistryMediumMCQMHT CET · 2026
The Arrhenius equation for the rate constant is $k = Ae^{-E_a/RT}$. $A$ chemical reaction will proceed more rapidly if there is a decrease in
A
$k$
B
$A$
C
$E_a$
D
$T$
Solution
(C) $1$. The rate of a chemical reaction is directly proportional to the rate constant $k$.
$2$. According to the Arrhenius equation $k = Ae^{-E_a/RT}$, the rate constant $k$ depends on the activation energy $E_a$ and temperature $T$.
$3$. As the activation energy $E_a$ decreases, the term $e^{-E_a/RT}$ increases, which leads to an increase in the value of $k$.
$4$. Therefore, a decrease in $E_a$ results in a faster reaction rate.
$5$. Thus, option $C$ is correct.
$2$. According to the Arrhenius equation $k = Ae^{-E_a/RT}$, the rate constant $k$ depends on the activation energy $E_a$ and temperature $T$.
$3$. As the activation energy $E_a$ decreases, the term $e^{-E_a/RT}$ increases, which leads to an increase in the value of $k$.
$4$. Therefore, a decrease in $E_a$ results in a faster reaction rate.
$5$. Thus, option $C$ is correct.
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336
ChemistryEasyMCQMHT CET · 2026
What is the term used for the minimum kinetic energy required for the reactant molecules to undergo a chemical reaction?
A
Potential energy
B
Bond energy
C
Thermal energy
D
Activation energy
Solution
(D) $1$. For a chemical reaction to occur, reactant molecules must collide with a certain minimum amount of kinetic energy.
$2$. This minimum energy threshold is known as the activation energy $(E_a)$.
$3$. Molecules with kinetic energy less than $E_a$ do not react upon collision, while those with energy equal to or greater than $E_a$ can successfully form products.
$2$. This minimum energy threshold is known as the activation energy $(E_a)$.
$3$. Molecules with kinetic energy less than $E_a$ do not react upon collision, while those with energy equal to or greater than $E_a$ can successfully form products.
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337
ChemistryMediumMCQMHT CET · 2026
What is the role of a catalyst in a chemical reaction?
A
To increase the activation energy of a reaction
B
To decrease the equilibrium constant of the reaction
C
To supply energy to the reactants
D
To decrease the activation energy of a reaction
Solution
(D) $1$. $A$ catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.
$2$. It functions by providing an alternative reaction pathway with a lower activation energy $(E_a)$.
$3$. By lowering the activation energy, a greater fraction of reactant molecules possess sufficient energy to cross the energy barrier, thereby increasing the reaction rate.
$4$. Therefore, the correct role is to decrease the activation energy of a reaction.
$2$. It functions by providing an alternative reaction pathway with a lower activation energy $(E_a)$.
$3$. By lowering the activation energy, a greater fraction of reactant molecules possess sufficient energy to cross the energy barrier, thereby increasing the reaction rate.
$4$. Therefore, the correct role is to decrease the activation energy of a reaction.
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338
ChemistryDifficultMCQMHT CET · 2026
The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $1/20^{th}$ of its initial value (in $\text{ s}$)?
A
$0.0529$
B
$0.0852$
C
$0.0499$
D
$0.0357$
Solution
(C) For a first order reaction, the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Given: $k = 60 \text{ s}^{-1}$ and $[R] = \frac{[R]_0}{20}$, so $\frac{[R]_0}{[R]} = 20$.
Substituting the values:
$60 = \frac{2.303}{t} \log(20)$
$t = \frac{2.303}{60} \times \log(20)$
Since $\log(20) = \log(2 \times 10) = \log(2) + \log(10) = 0.3010 + 1 = 1.3010$.
$t = \frac{2.303 \times 1.3010}{60}$
$t = \frac{2.9962}{60} \approx 0.0499 \text{ s}$.
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Given: $k = 60 \text{ s}^{-1}$ and $[R] = \frac{[R]_0}{20}$, so $\frac{[R]_0}{[R]} = 20$.
Substituting the values:
$60 = \frac{2.303}{t} \log(20)$
$t = \frac{2.303}{60} \times \log(20)$
Since $\log(20) = \log(2 \times 10) = \log(2) + \log(10) = 0.3010 + 1 = 1.3010$.
$t = \frac{2.303 \times 1.3010}{60}$
$t = \frac{2.9962}{60} \approx 0.0499 \text{ s}$.
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339
ChemistryEasyMCQMHT CET · 2026
What is the order of radioactive decay of a substance?
A
$0$
B
$1$
C
$5$
D
$2$
Solution
(B) Radioactive decay is a process where an unstable atomic nucleus loses energy by radiation.
It is experimentally observed that the rate of radioactive decay is directly proportional to the number of radioactive nuclei present at that time.
Mathematically, $-\frac{dN}{dt} = \lambda N$, where $\lambda$ is the decay constant.
Since the rate depends on the first power of the concentration of the reactant (nuclei), it follows first-order kinetics.
Therefore, the order of radioactive decay is $1$.
It is experimentally observed that the rate of radioactive decay is directly proportional to the number of radioactive nuclei present at that time.
Mathematically, $-\frac{dN}{dt} = \lambda N$, where $\lambda$ is the decay constant.
Since the rate depends on the first power of the concentration of the reactant (nuclei), it follows first-order kinetics.
Therefore, the order of radioactive decay is $1$.
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340
ChemistryDifficultMCQMHT CET · 2026
$A$ first order reaction is $60\%$ complete in $20 \text{ minutes}$. How long will the reaction take to complete $84\%$ (in $\text{ min}$)?
A
$54$
B
$68$
C
$40$
D
$76$
Solution
(C) For a first order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $60\%$ completion, $[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 20 \text{ min}$.
$k = \frac{2.303}{20} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{20} \log(2.5) \approx \frac{2.303 \times 0.3979}{20} \approx 0.0458 \text{ min}^{-1}$.
For $84\%$ completion, $[A]_t = [A]_0 - 0.84[A]_0 = 0.16[A]_0$.
$t = \frac{2.303}{k} \log \frac{[A]_0}{0.16[A]_0} = \frac{2.303}{0.0458} \log(6.25)$.
Since $\log(6.25) = 2 \log(2.5) \approx 2 \times 0.3979 = 0.7958$.
$t = \frac{2.303 \times 0.7958}{0.0458} \approx 40 \text{ min}$.
For $60\%$ completion, $[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 20 \text{ min}$.
$k = \frac{2.303}{20} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{20} \log(2.5) \approx \frac{2.303 \times 0.3979}{20} \approx 0.0458 \text{ min}^{-1}$.
For $84\%$ completion, $[A]_t = [A]_0 - 0.84[A]_0 = 0.16[A]_0$.
$t = \frac{2.303}{k} \log \frac{[A]_0}{0.16[A]_0} = \frac{2.303}{0.0458} \log(6.25)$.
Since $\log(6.25) = 2 \log(2.5) \approx 2 \times 0.3979 = 0.7958$.
$t = \frac{2.303 \times 0.7958}{0.0458} \approx 40 \text{ min}$.
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341
ChemistryDifficultMCQMHT CET · 2026
The half-life period of a first-order reaction is $1386 \text{ s}$. The rate constant of the reaction is:
A
$5.5 \times 10^{-2} \text{ s}^{-1}$
B
$0.5 \times 10^{-3} \text{ s}^{-1}$
C
$5 \times 10^{-2} \text{ s}^{-1}$
D
$5 \times 10^{-3} \text{ s}^{-1}$
Solution
(B) For a first-order reaction, the relationship between the rate constant $(k)$ and the half-life $(t_{1/2})$ is given by:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1386 \text{ s}$.
Substituting the value:
$k = \frac{0.693}{1386} \text{ s}^{-1}$
$k = \frac{693 \times 10^{-3}}{1386} \text{ s}^{-1}$
$k = 0.5 \times 10^{-3} \text{ s}^{-1}$
Thus, the rate constant is $5 \times 10^{-4} \text{ s}^{-1}$ or $0.5 \times 10^{-3} \text{ s}^{-1}$.
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1386 \text{ s}$.
Substituting the value:
$k = \frac{0.693}{1386} \text{ s}^{-1}$
$k = \frac{693 \times 10^{-3}}{1386} \text{ s}^{-1}$
$k = 0.5 \times 10^{-3} \text{ s}^{-1}$
Thus, the rate constant is $5 \times 10^{-4} \text{ s}^{-1}$ or $0.5 \times 10^{-3} \text{ s}^{-1}$.
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342
ChemistryDifficultMCQMHT CET · 2026
The half-life of a first-order reaction is $850 \text{ s}$. The initial concentration of the reactant is $[R]_0 = 0.06 \text{ mol dm}^{-3}$. What concentration $[R]$ would remain after $t = 1200 \text{ s}$?
A
$0.023 \text{ mol dm}^{-3}$
B
$0.035 \text{ mol dm}^{-3}$
C
$5.25 \text{ mol dm}^{-3}$
D
$6.25 \text{ mol dm}^{-3}$
Solution
(A) Step $1$: Calculate the rate constant $k$ for the first-order reaction using $k = \frac{0.693}{t_{1/2}}$.
$k = \frac{0.693}{850 \text{ s}} \approx 8.153 \times 10^{-4} \text{ s}^{-1}$.
Step $2$: Use the integrated rate law for a first-order reaction: $[R] = [R]_0 e^{-kt}$.
$[R] = 0.06 \times e^{-(8.153 \times 10^{-4} \times 1200)}$.
Step $3$: Calculate the exponent: $-(8.153 \times 10^{-4} \times 1200) \approx -0.97836$.
Step $4$: Calculate the final concentration: $[R] = 0.06 \times e^{-0.97836} \approx 0.06 \times 0.3759 \approx 0.02255 \text{ mol dm}^{-3}$.
Rounding to two significant figures, we get $0.023 \text{ mol dm}^{-3}$. Thus, option $A$ is correct.
$k = \frac{0.693}{850 \text{ s}} \approx 8.153 \times 10^{-4} \text{ s}^{-1}$.
Step $2$: Use the integrated rate law for a first-order reaction: $[R] = [R]_0 e^{-kt}$.
$[R] = 0.06 \times e^{-(8.153 \times 10^{-4} \times 1200)}$.
Step $3$: Calculate the exponent: $-(8.153 \times 10^{-4} \times 1200) \approx -0.97836$.
Step $4$: Calculate the final concentration: $[R] = 0.06 \times e^{-0.97836} \approx 0.06 \times 0.3759 \approx 0.02255 \text{ mol dm}^{-3}$.
Rounding to two significant figures, we get $0.023 \text{ mol dm}^{-3}$. Thus, option $A$ is correct.
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343
ChemistryDifficultMCQMHT CET · 2026
For a first-order reaction, $20 \%$ of the initial concentration remains after $10 \text{ min}$. What is the rate constant of the reaction (in $\text{ min}^{-1}$)? (Given: $\log_{10}(5) = 0.6989$)
A
$1.609$
B
$6.989$
C
$16.09$
D
$0.1609$
Solution
(D) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$.
Given: $[A]_t = 20 \% \text{ of } [A]_0 = 0.2 [A]_0$, so $\frac{[A]_0}{[A]_t} = \frac{1}{0.2} = 5$.
Given $t = 10 \text{ min}$ and $\log_{10}(5) = 0.6989$.
Substituting the values: $k = \frac{2.303}{10} \times 0.6989$.
$k = 0.2303 \times 0.6989 \approx 0.1609 \text{ min}^{-1}$.
Given: $[A]_t = 20 \% \text{ of } [A]_0 = 0.2 [A]_0$, so $\frac{[A]_0}{[A]_t} = \frac{1}{0.2} = 5$.
Given $t = 10 \text{ min}$ and $\log_{10}(5) = 0.6989$.
Substituting the values: $k = \frac{2.303}{10} \times 0.6989$.
$k = 0.2303 \times 0.6989 \approx 0.1609 \text{ min}^{-1}$.
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344
ChemistryDifficultMCQMHT CET · 2026
The half-life of a first-order reaction is $20 \text{ minutes}$. What is the rate constant (in $\text{min}^{-1}$) for this reaction?
A
$0.0347$
B
$0.5$
C
$13.86$
D
$34.67$
Solution
(A) For a first-order reaction, the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is given by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 20 \text{ min}$.
Substituting the value into the formula:
$k = \frac{0.693}{20 \text{ min}}$
$k = 0.03465 \text{ min}^{-1}$
Rounding to four decimal places, we get $k \approx 0.0347 \text{ min}^{-1}$.
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 20 \text{ min}$.
Substituting the value into the formula:
$k = \frac{0.693}{20 \text{ min}}$
$k = 0.03465 \text{ min}^{-1}$
Rounding to four decimal places, we get $k \approx 0.0347 \text{ min}^{-1}$.
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345
ChemistryDifficultMCQMHT CET · 2026
$A$ first-order reaction takes $16 \text{ minutes}$ for its $50\%$ completion. What fraction of the reactant would react in $32 \text{ minutes}$ from the beginning?
A
$1/2$
B
$1/4$
C
$1/8$
D
$3/4$
Solution
(D) For a first-order reaction, the half-life $t_{1/2}$ is given as $16 \text{ minutes}$.
In $32 \text{ minutes}$, the number of half-lives elapsed is $n = \frac{32}{16} = 2$.
The fraction of reactant remaining after $n$ half-lives is given by $(\frac{1}{2})^n$.
Remaining fraction $= (\frac{1}{2})^2 = \frac{1}{4}$.
The fraction that has reacted is $1 - \text{remaining fraction} = 1 - \frac{1}{4} = \frac{3}{4}$.
In $32 \text{ minutes}$, the number of half-lives elapsed is $n = \frac{32}{16} = 2$.
The fraction of reactant remaining after $n$ half-lives is given by $(\frac{1}{2})^n$.
Remaining fraction $= (\frac{1}{2})^2 = \frac{1}{4}$.
The fraction that has reacted is $1 - \text{remaining fraction} = 1 - \frac{1}{4} = \frac{3}{4}$.
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346
ChemistryDifficultMCQMHT CET · 2026
For the reaction, $A + B \rightarrow P$, the rate law is $\text{rate} = k[A][B]^2$. The rate of reaction is $0.25 \text{ Ms}^{-1}$ when $[A] = 1 \text{ M}$ and $[B] = 0.2 \text{ M}$ at $25^\circ \text{C}$. Calculate the rate constant $k$ of the reaction at the same temperature.
A
$6.25 \text{ M}^{-2}\text{s}^{-1}$
B
$0.25 \text{ M}^{-2}\text{s}^{-1}$
C
$5.0 \text{ M}^{-2}\text{s}^{-1}$
D
$75.0 \text{ M}^{-2}\text{s}^{-1}$
Solution
(A) Given rate law: $\text{rate} = k[A][B]^2$
Substitute the given values: $0.25 \text{ Ms}^{-1} = k(1 \text{ M})(0.2 \text{ M})^2$
$0.25 = k(1)(0.04)$
$k = \frac{0.25}{0.04}$
$k = 6.25 \text{ M}^{-2}\text{s}^{-1}$
Substitute the given values: $0.25 \text{ Ms}^{-1} = k(1 \text{ M})(0.2 \text{ M})^2$
$0.25 = k(1)(0.04)$
$k = \frac{0.25}{0.04}$
$k = 6.25 \text{ M}^{-2}\text{s}^{-1}$
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347
ChemistryMediumMCQMHT CET · 2026
The rate constant of a reaction is $k = 3.28 \times 10^{-4} \text{ s}^{-1}$. Find the order of the reaction.
A
Zero order
B
First order
C
Second order
D
Third order
Solution
(B) Step $1$: Identify the unit of the rate constant $k$. The given unit is $\text{s}^{-1}$.
Step $2$: Recall the general formula for the units of the rate constant for a reaction of order $n$: $\text{unit} = (\text{concentration})^{1-n} \times \text{time}^{-1}$.
Step $3$: For a first-order reaction $(n=1)$, the unit is $(\text{mol L}^{-1})^{1-1} \times \text{s}^{-1} = \text{s}^{-1}$.
Step $4$: Since the given unit $\text{s}^{-1}$ matches the unit of a first-order reaction, the reaction is of the first order.
Step $2$: Recall the general formula for the units of the rate constant for a reaction of order $n$: $\text{unit} = (\text{concentration})^{1-n} \times \text{time}^{-1}$.
Step $3$: For a first-order reaction $(n=1)$, the unit is $(\text{mol L}^{-1})^{1-1} \times \text{s}^{-1} = \text{s}^{-1}$.
Step $4$: Since the given unit $\text{s}^{-1}$ matches the unit of a first-order reaction, the reaction is of the first order.
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348
ChemistryDifficultMCQMHT CET · 2026
The rate of the reaction $2A + 3B \rightarrow 2C + D$ is $6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$, when $[A] = [B] = 0.3 \text{ mol dm}^{-3}$. If the reaction is of first order with respect to $A$ and zeroth order with respect to $B$, find the rate constant $k$.
A
$1 \times 10^{-3} \text{ s}^{-1}$
B
$2 \times 10^{-3} \text{ s}^{-1}$
C
$3 \times 10^{-3} \text{ s}^{-1}$
D
$4 \times 10^{-3} \text{ s}^{-1}$
Solution
(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^0 = k[A]$.
Given: $\text{Rate} = 6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $6 \times 10^{-4} = k \times 0.3$.
Solving for $k$: $k = \frac{6 \times 10^{-4}}{0.3} = 20 \times 10^{-4} \text{ s}^{-1} = 2 \times 10^{-3} \text{ s}^{-1}$.
Given: $\text{Rate} = 6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $6 \times 10^{-4} = k \times 0.3$.
Solving for $k$: $k = \frac{6 \times 10^{-4}}{0.3} = 20 \times 10^{-4} \text{ s}^{-1} = 2 \times 10^{-3} \text{ s}^{-1}$.
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349
ChemistryDifficultMCQMHT CET · 2026
The rate of reaction $A + B \rightarrow P$ is $4 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}$. When $[A] = 0.2 \text{ mol dm}^{-3}$ and $[B] = 0.1 \text{ mol dm}^{-3}$, what is the rate constant of the reaction, if it is first order with respect to $A$ and second order with respect to $B$?
A
$10 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$
B
$20 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$
C
$25 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$
D
$40 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$
Solution
(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^2$.
Given: $\text{Rate} = 4 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}$, $[A] = 0.2 \text{ mol dm}^{-3}$, $[B] = 0.1 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $4 \times 10^{-2} = k(0.2)^1(0.1)^2$.
$4 \times 10^{-2} = k(0.2)(0.01)$.
$4 \times 10^{-2} = k(0.002)$.
$k = \frac{4 \times 10^{-2}}{2 \times 10^{-3}} = 2 \times 10^1 = 20 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$.
Given: $\text{Rate} = 4 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}$, $[A] = 0.2 \text{ mol dm}^{-3}$, $[B] = 0.1 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $4 \times 10^{-2} = k(0.2)^1(0.1)^2$.
$4 \times 10^{-2} = k(0.2)(0.01)$.
$4 \times 10^{-2} = k(0.002)$.
$k = \frac{4 \times 10^{-2}}{2 \times 10^{-3}} = 2 \times 10^1 = 20 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$.
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350
ChemistryDifficultMCQMHT CET · 2026
The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:
A
$0.4$
B
$0.8$
C
$1.2$
D
$2.0$
Solution
(B) For a first-order reaction, the rate law is given by: $\text{Rate} = k[N_2O_5]$.
Given: $\text{Rate} = 2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$ and $k = 3.0 \times 10^{-5} \text{ s}^{-1}$.
Substituting the values into the rate equation: $2.4 \times 10^{-5} = (3.0 \times 10^{-5}) \times [N_2O_5]$.
Solving for $[N_2O_5]$: $[N_2O_5] = \frac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$.
$[N_2O_5] = 0.8 \text{ mol L}^{-1}$.
Given: $\text{Rate} = 2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$ and $k = 3.0 \times 10^{-5} \text{ s}^{-1}$.
Substituting the values into the rate equation: $2.4 \times 10^{-5} = (3.0 \times 10^{-5}) \times [N_2O_5]$.
Solving for $[N_2O_5]$: $[N_2O_5] = \frac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$.
$[N_2O_5] = 0.8 \text{ mol L}^{-1}$.
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