MHT CET 2026 Chemistry Question Paper with Answer and Solution

(D) The oxidizing power of perhalate ions depends on the ease of reduction of the central halogen atom.
$ClO_4^-$ is a weak oxidizing agent because the $Cl$ atom is sterically protected by four oxygen atoms and is in its highest oxidation state.
$BrO_4^-$ is a strong oxidizing agent because $Br-O$ bonds are weaker than $Cl-O$ bonds, making it easier to reduce.
$IO_4^-$ is a moderate oxidizing agent, weaker than $BrO_4^-$ but stronger than $ClO_4^-$.
Thus, the decreasing order of oxidizing power is $BrO_4^- > IO_4^- > ClO_4^-$.

(A) The acidic strength of oxyacids of chlorine increases with an increase in the oxidation state of the central chlorine atom.
$1$. Oxidation state of $Cl$ in $HClO$ is $+1$.
$2$. Oxidation state of $Cl$ in $HClO_2$ is $+3$.
$3$. Oxidation state of $Cl$ in $HClO_3$ is $+5$.
$4$. Oxidation state of $Cl$ in $HClO_4$ is $+7$.
As the oxidation state increases, the electron-withdrawing power of the chlorine atom increases, which weakens the $O-H$ bond and facilitates the release of $H^+$ ions.
Therefore, $HClO$ (Hypochlorous acid) is the weakest acid.

(B) $1$. Thermal stability of interhalogen compounds depends on the difference in electronegativity between the two halogen atoms.
$2$. $A$ larger difference in electronegativity leads to a stronger polar bond, resulting in higher thermal stability.
$3$. The electronegativity values are: $F = 4.0$, $Cl = 3.0$, $Br = 2.8$, $I = 2.5$.
$4$. Calculating the electronegativity difference $(\Delta EN)$:
- For $IBr$: $|2.5 - 2.8| = 0.3$
- For $BrCl$: $|2.8 - 3.0| = 0.2$
- For $BrF$: $|2.8 - 4.0| = 1.2$
- For $ClF$: $|3.0 - 4.0| = 1.0$
$5$. Since $BrCl$ has the smallest electronegativity difference $(0.2)$, it has the weakest bond and therefore the lowest thermal stability.

(B) Let the oxidation state of chlorine be $x$.
In $HClO$: $1 + x + (-2) = 0 \implies x = +1$.
In $HClO_2$: $1 + x + 2(-2) = 0 \implies 1 + x - 4 = 0 \implies x = +3$.
In $HClO_3$: $1 + x + 3(-2) = 0 \implies 1 + x - 6 = 0 \implies x = +5$.
In $HClO_4$: $1 + x + 4(-2) = 0 \implies 1 + x - 8 = 0 \implies x = +7$.
Thus, $HClO_2$ exhibits the $+3$ oxidation state.

(C) $1$. The bond dissociation enthalpy generally decreases down the group due to an increase in atomic size and bond length.
$2$. However, $F_2$ has an exceptionally low bond dissociation enthalpy compared to $Cl_2$ and $Br_2$.
$3$. This is due to the small size of the $F$ atom, which leads to strong inter-electronic repulsions between the non-bonding electrons of the two $F$ atoms in the $F-F$ bond.
$4$. Therefore, the correct decreasing order is $Cl_2 > Br_2 > F_2 > I_2$.

(B) Step $1$: The contact process involves the oxidation of $SO_2$ to $SO_3$ as a key step: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$.
Step $2$: This reaction is exothermic and reversible. To increase the rate of reaction, a catalyst is required.
Step $3$: Historically, platinised asbestos was used, but it is easily poisoned by impurities. Currently, $V_2O_5$ (Vanadium pentoxide) is used as the standard industrial catalyst for this process.
Step $4$: Since the original options provided were incorrect, $V_2O_5$ is the correct catalyst.

(C) $1$. Ozone $(O_3)$ acts as a strong oxidizing agent because it readily decomposes to give nascent oxygen $(O_3 \rightarrow O_2 + [O])$.
$2$. In option $(C)$, $H_2O_2$ is oxidized to $O_2$ by $O_3$, while $O_3$ itself is reduced to $O_2$. This reaction demonstrates the reducing nature of ozone.
$3$. In options $(A)$, $(B)$, and $(D)$, ozone acts as an oxidizing agent by oxidizing $NO$ to $NO_2$, $I^-$ to $I_2$, and $PbS$ to $PbSO_4$ respectively.

(D) $1$. In the Haber process, the reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
$2$. Finely divided iron $(Fe)$ acts as the catalyst for this reaction.
$3$. Molybdenum $(Mo)$ or a mixture of oxides like $Al_2O_3$ and $K_2O$ is added to increase the efficiency of the iron catalyst.
$4$. $A$ substance that increases the activity of a catalyst is called a promoter.
$5$. Therefore, $Al_2O_3$ acts as a promoter.

(D) The lanthanoids are a series of elements from atomic number $57$ to $71$. Among these, $Pm$ (Promethium, atomic number $61$) is the only synthetic radioactive element. All its isotopes are radioactive.
Therefore, the correct option is $D$.

(C) In optical fiber communication systems, Erbium-doped fiber amplifiers $(EDFAs)$ are widely used. Erbium ions $(Er^{3+})$ are doped into the silica fiber core because they provide optical amplification in the $1550 \text{ nm}$ wavelength window, which corresponds to the minimum attenuation region of silica optical fibers.

(A) $1$. Lanthanide contraction is the steady decrease in the atomic and ionic radii of lanthanides with an increase in atomic number.
$2$. As we move from $La$ to $Lu$, electrons are added to the $4f$ subshell.
$3$. The $4f$ orbitals have a very diffuse shape, which results in poor shielding of the nuclear charge by the $4f$ electrons.
$4$. Due to this poor shielding, the effective nuclear charge experienced by the outer electrons increases, causing the electron cloud to be pulled closer to the nucleus, resulting in a decrease in atomic size.
$5$. Therefore, the main cause is the poor shielding of one $4f$ electron by another.

(C) In actinoids, the $5f$, $6d$, and $7s$ energy levels are very close to each other in energy. Due to this comparable energy, electrons from all these subshells can participate in bonding, which allows actinoids to exhibit a greater range of oxidation states compared to lanthanoids.

(B) The ground state electronic configuration of the given lanthanoids are:
$(1)$ $La (Z=57): [Xe] 5d^1 6s^2$
$(2)$ $Yb (Z=70): [Xe] 4f^{14} 6s^2$
$(3)$ $Eu (Z=63): [Xe] 4f^7 6s^2$
$(4)$ $Gd (Z=64): [Xe] 4f^7 5d^1 6s^2$
Among these, $Yb$ has a completely filled $4f$ orbital $(4f^{14})$.
Therefore, the correct option is $B$.

(C) The actinoid series consists of elements with atomic numbers from $Z = 89$ to $Z = 103$. The first element following the actinoid series is known as the first post-actinoid element.
Atomic number of the last actinoid $(Lawrencium)$ is $103$.
Therefore, the atomic number of the first post-actinoid element is $103 + 1 = 104$ $(Rutherfordium)$.
Thus, the correct option is $C$.

(D) $1$. In the actinoid series, as the atomic number increases, the ionic radius decreases due to actinoid contraction.
$2$. Actinoid contraction is similar to lanthanoid contraction, caused by the poor shielding effect of $5f$ electrons.
$3$. Among the given options, $Lr$ (Lawrencium, $Z=103$) has the highest atomic number.
$4$. Therefore, $Lr^{3+}$ has the smallest ionic size.

(B) The general electronic configuration of lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
In the $+3$ oxidation state, the lanthanoids lose two $6s$ electrons and one $4f$ or $5d$ electron.
For $Lu$ (Lutetium, atomic number $71$), the ground state configuration is $[Xe] 4f^{14} 5d^1 6s^2$.
In the $+3$ oxidation state, $Lu$ loses two $6s$ electrons and one $5d$ electron, resulting in the configuration $[Xe] 4f^{14}$.
Thus, $Lu^{3+}$ has a stable $f^{14}$ configuration.

(A) Inner transition elements consist of the lanthanoids ($Z = 58$ to $71$) and the actinoids ($Z = 90$ to $103$).
$(1)$ $Ac$ $(Z = 89)$ is a transition element (d-block).
$(2)$ $Lu$ $(Z = 71)$ is the last element of the lanthanoid series.
$(3)$ $Ho$ $(Z = 67)$ is a lanthanoid.
$(4)$ $Rf$ $(Z = 104)$ is a transition element (d-block) belonging to group $4$.
Therefore, $Rf$ is not an inner transition element.

(B) The electronic configuration of zinc ($Zn$, atomic number $30$) is $[Ar] 3d^{10} 4s^2$.
In $Zn^{2+}$ ion, the two electrons from the $4s$ orbital are removed, resulting in the configuration $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled, it is highly stable and does not allow for the loss of further electrons to show variable oxidation states.
Therefore, zinc only exhibits a $+2$ oxidation state.

(C) An ion is paramagnetic if it contains at least one unpaired electron.
$(1)$ $Ti^{+2}$ $([Ar] 3d^2)$: $2$ unpaired electrons (Paramagnetic).
$(2)$ $Cu^{+2}$ $([Ar] 3d^9)$: $1$ unpaired electron (Paramagnetic).
$(3)$ $Mn^{+3}$ $([Ar] 3d^4)$: $4$ unpaired electrons (Paramagnetic).
Since all ions in option $(C)$ have unpaired electrons, they are all paramagnetic.

(C) Step $1$: Identify the position of $Co$ (Cobalt) in the periodic table. $Co$ has an atomic number of $27$ and belongs to the $3d$ transition series.
Step $2$: Identify the position of $Mo$ (Molybdenum) in the periodic table. $Mo$ has an atomic number of $42$ and belongs to the $4d$ transition series.
Step $3$: Therefore, $Co$ and $Mo$ belong to the $3d$ and $4d$ series respectively.

(D) $1$. $A$ transition metal ion forms colored compounds if it has partially filled $d$-orbitals (i.e., $d^1$ to $d^9$ configuration) due to $d-d$ transitions.
$2$. Electronic configurations:
- $Sc^{3+}$ $(Z=21)$: $[Ar] 3d^0$ (no $d$-electrons, colorless).
- $Ti^{4+}$ $(Z=22)$: $[Ar] 3d^0$ (no $d$-electrons, colorless).
- $Cu^{+}$ $(Z=29)$: $[Ar] 3d^{10}$ (fully filled $d$-orbitals, colorless).
- $V^{3+}$ $(Z=23)$: $[Ar] 3d^2$ (partially filled $d$-orbitals, colored).
$3$. Therefore, $V^{3+}$ forms colored compounds.

(C) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$, where $n$ is the number of unpaired electrons.
$(A)$ $Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$, $n=1$.
$(B)$ $Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$, $n=3$.
$(C)$ $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$, $n=5$.
$(D)$ $Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$, $n=5$.
$(E)$ $Sc^{3+}$ $(Z=21)$: $[Ar] 3d^0$, $n=0$.
Since $Mn^{2+}$ and $Fe^{3+}$ both have $n=5$ unpaired electrons, they exhibit the same spin-only magnetic moment. Thus, the correct option is $(C)$.

(C) $1$. The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
$2$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
$3$. In $3d^9$, there is $1$ unpaired electron $(n = 1)$.
$4$. The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$.
$5$. Substituting $n = 1$: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(A) $1$. The colour of transition metal ions depends on the presence of unpaired $d$-electrons, which allow for $d-d$ transitions.
$2$. The electronic configuration of $Sc$ $(Z=21)$ is $[Ar] 3d^1 4s^2$. In the $+3$ oxidation state, it becomes $Sc^{3+} = [Ar] 3d^0$.
$3$. Since $Sc^{3+}$ has no unpaired electrons ($d^0$ configuration), $d-d$ transitions are not possible, making its compounds colourless.
$4$. $Ni^{3+}$, $Cr^{3+}$, and $V^{3+}$ have partially filled $d$-orbitals ($d^7$, $d^3$, and $d^2$ respectively), which allow for $d-d$ transitions, resulting in coloured compounds.

(A) $1$. Transition metal compounds are often coloured due to the presence of unpaired $d$-electrons.
$2$. When light falls on these compounds, the $d$-electrons absorb specific wavelengths of visible light to jump from a lower energy $d$-orbital to a higher energy $d$-orbital.
$3$. This process is known as a $d-d$ transition.
$4$. The light that is not absorbed is transmitted or reflected, which gives the compound its characteristic colour.

(B) $1$. The general electronic configuration of transition elements is $(n-1)d^{1-10} ns^{0-2}$.
$2$. For the $3d$ series, $n=4$, so the configuration is $[Ar] 3d^{1-10} 4s^{1-2}$.
$3$. For the $4d$ series, $n=5$, so the configuration is $[Kr] 4d^{1-10} 5s^{0-2}$.
$4$. For the $5d$ series, $n=6$, so the configuration is $[Xe] 4f^{14} 5d^{1-10} 6s^{0-2}$.
$5$. Therefore, the configuration $[Kr] 4d^{1-10} 5s^{0-2}$ corresponds to the $4d$ series.

(A) Transition metals have a crystal lattice structure with interstitial sites (voids) of appropriate size. Small atoms like $H$, $C$, $N$, and $O$ can easily occupy these voids, forming interstitial compounds. Therefore, the correct reason is the presence of these interstitial sites.

(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$, where $n$ is the number of unpaired electrons.
$1$. $Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$, $n=1$, $\mu = \sqrt{1(3)} = 1.73 \text{ B.M.}$
$2$. $Zn^{2+}$ $(Z=30)$: $[Ar] 3d^{10}$, $n=0$, $\mu = 0 \text{ B.M.}$
$3$. $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$, $n=4$, $\mu = \sqrt{4(6)} = 4.90 \text{ B.M.}$
$4$. $Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$, $n=4$, $\mu = \sqrt{4(6)} = 4.90 \text{ B.M.}$
$5$. $Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$, $n=3$, $\mu = \sqrt{3(5)} = 3.87 \text{ B.M.}$
$6$. $Cu^{2+}$ $(Z=29)$: $[Ar] 3d^9$, $n=1$, $\mu = 1.73 \text{ B.M.}$
Comparing the values, $Cr^{2+}$ and $Fe^{2+}$ both have $n=4$ and thus the same magnetic moment.

(B) The $d$-block elements are arranged in four transition series based on the filling of $d$-orbitals:
$1$. The $3d$-series (first transition series) involves the filling of $3d$-orbitals.
$2$. The $4d$-series (second transition series) involves the filling of $4d$-orbitals.
$3$. The $5d$-series (third transition series) involves the filling of $5d$-orbitals.
$4$. The $6d$-series (fourth transition series) involves the filling of $6d$-orbitals.
Therefore, there are a total of $4$ transition series.

(C) Step $1$: Determine the electronic configuration and number of unpaired electrons for each ion.
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$. Unpaired electrons = $4$.
$Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$. Unpaired electrons = $5$.
$Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$. Unpaired electrons = $3$.
$Ni^{2+}$ $(Z=28)$: $[Ar] 3d^8$. Unpaired electrons = $2$.
$Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$. Unpaired electrons = $4$.
Step $2$: Compare the number of unpaired electrons.
$Fe^{2+}$ has $4$ unpaired electrons and $Cr^{2+}$ has $4$ unpaired electrons.
Therefore, the pair $Fe^{2+}$ and $Cr^{2+}$ has the same number of unpaired electrons.

(D) $1$. Manganese $(Mn)$ has the atomic number $25$ and its electronic configuration is $[Ar] 3d^5 4s^2$.
$2$. It can lose all $7$ valence electrons ($5$ from $3d$ and $2$ from $4s$) to achieve a stable configuration.
$3$. Therefore, the highest oxidation state exhibited by manganese is $+7$, as seen in compounds like potassium permanganate $(KMnO_4)$.

(D) Brass is an alloy primarily composed of copper $(Cu)$ and zinc $(Zn)$. Therefore, the correct option is $D$.

(B) $1$. The $2^{nd}$ transition series corresponds to the $4d$ series, which starts from Yttrium $(Z=39)$ and ends at Cadmium $(Z=48)$.
$2$. These elements follow the noble gas core of Krypton ($[Kr]$, $Z=36$).
$3$. The general valence shell electronic configuration for the $d$-block elements is $(n-1)d^{1-10} ns^{1-2}$.
$4$. For the $2^{nd}$ transition series, $n=5$. Therefore, the configuration is $[Kr] 4d^{1-10} 5s^{0-2}$ (where $5s$ can have $0, 1,$ or $2$ electrons depending on the element, such as $Pd$ which is $4d^{10} 5s^0$).

(C) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any one of its oxidation states.
$Sc$ $([Ar] 3d^1 4s^2)$ has a partially filled $d$-orbital.
$Cu$ $([Ar] 3d^{10} 4s^1)$ forms $Cu^{2+}$ $([Ar] 3d^9)$, which has a partially filled $d$-orbital.
$Au$ $([Xe] 4f^{14} 5d^{10} 6s^1)$ forms $Au^{3+}$ $([Xe] 4f^{14} 5d^8)$, which has a partially filled $d$-orbital.
$Cd$ $([Kr] 4d^{10} 5s^2)$ has a completely filled $d$-orbital in its ground state and in its only common oxidation state $Cd^{2+}$ $([Kr] 4d^{10})$.
Therefore, $Cd$ is not considered a transition element.

(D) Step $1$: Calculate the oxidation state of the metal in each species.
Step $2$: For $MnO_4^{-}$, let the oxidation state of $Mn$ be $x$. $x + 4(-2) = -1 \implies x = +7$.
Step $3$: For $[Cr(CN)_6]^{3-}$, let the oxidation state of $Cr$ be $x$. $x + 6(-1) = -3 \implies x = +3$.
Step $4$: For $Cr_2O_3$, let the oxidation state of $Cr$ be $x$. $2x + 3(-2) = 0 \implies 2x = 6 \implies x = +3$.
Step $5$: For $CrO_2Cl_2$, let the oxidation state of $Cr$ be $x$. $x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6$.
Step $6$: Thus, $CrO_2Cl_2$ contains the metal in the $+6$ oxidation state.

(C) The $Fischer-Tropsch$ process is a collection of chemical reactions that converts a mixture of carbon monoxide and hydrogen into liquid hydrocarbons. The catalysts typically used in this process are metals such as cobalt, iron, or ruthenium. Specifically, a mixture of cobalt and thorium $(Co-Th)$ is a well-known catalyst used for this synthesis. Therefore, option $C$ is correct.

(A) The melting points of $3d$ transition elements depend on the number of unpaired electrons available for metallic bonding. $Cr$ (Chromium) has the electronic configuration $[Ar] 3d^5 4s^1$. It has $6$ unpaired electrons, which leads to strong metallic bonding and consequently the highest melting point among the given options $(1907 \text{ °C})$.

(A) Step $1$: Transition metal ions exhibit colour due to $d-d$ electronic transitions.
Step $2$: For a $d-d$ transition to occur, the $d$-orbital must be partially filled, meaning there must be at least one unpaired electron.
Step $3$: If a transition metal ion is colourless, it implies that no $d-d$ transition is possible.
Step $4$: This occurs when the $d$-orbital is either completely empty $(d^0)$ or completely filled $(d^{10})$, both of which correspond to $0$ unpaired electrons.

(D) $1$. Transition elements are defined as elements that have a partially filled $d$-orbital in their ground state or in any of their common oxidation states.
$2$. $Mn$ $(Z=25)$, $Ag$ $(Z=47)$, $Zr$ $(Z=40)$, $Au$ $(Z=79)$, $Mo$ $(Z=42)$, and $Pt$ $(Z=78)$ are all transition elements.
$3$. $Sn$ $(Z=50)$ is a post-transition metal in group $14$ ($p$-block).
$4$. $Pm$ $(Z=61)$ is a lanthanoid (inner transition element).
$5$. Therefore, the pair $(Sn, Pm)$ does not contain any transition elements.

(B) $1$. The color observed by the human eye is the complementary color of the light absorbed by the substance.
$2$. According to the color wheel, the complementary color of $red$ is $green$ or $blue-green$.
$3$. Among the given options, $blue$ is the closest complementary color to $red$ in the context of color absorption theory.

(A) The magnitude of crystal field splitting energy $\Delta_0$ depends on:
$1$. Oxidation state of the metal ion: Higher oxidation state leads to higher $\Delta_0$.
$2$. Nature of the ligand: Strong field ligands (like $CN^-$) result in higher $\Delta_0$ than weak field ligands (like $H_2O$).
$3$. Principal quantum number of the metal: $4d$ and $5d$ series have higher $\Delta_0$ than $3d$ series.
Comparing the complexes:
- For $I$ $(Co^{2+})$ and $II$ $(Co^{3+})$, both have $H_2O$ ligands. Since $Co^{3+}$ has a higher oxidation state, $\Delta_0(II) > \Delta_0(I)$.
- For $II$ $(Co^{3+})$ and $III$ $(Fe^{3+})$, both have $H_2O$ ligands. $Co^{3+}$ has a higher effective nuclear charge than $Fe^{3+}$, so $\Delta_0(II) > \Delta_0(III)$.
- For $II$ ($H_2O$ ligand) and $IV$ ($CN^-$ ligand), $CN^-$ is a much stronger field ligand than $H_2O$, so $\Delta_0(IV) > \Delta_0(II)$.
Thus, the correct comparison is $I < II$.

(A) According to the spectrochemical series, the order of field strength of ligands is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < NH_3 < en < NO_2^{-} < CN^{-} < CO$.
Stronger ligands cause greater splitting of $d$-orbitals $(\Delta_o)$.
Among the given options, $CN^{-}$ is the strongest ligand, hence it causes the maximum splitting of $d$-orbitals.

(B) $1$. For $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand, causing pairing. Unpaired electrons $n = 0$.
$2$. For $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand, no pairing. $n = 4$.
$3$. For $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand, no pairing. $n = 2$.
$4$. For $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand, causing pairing. $n = 0$.
Comparing the values, $[CoF_6]^{3-}$ has the highest number of unpaired electrons $(n = 4)$.

(A) According to the spectrochemical series, the order of field strength for the given ligands is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Comparing the given options ($Br^{-}$, $OH^{-}$, $en$, $H_2O$), $Br^{-}$ occupies the position with the lowest field strength in the spectrochemical series.
Therefore, $Br^{-}$ is the weakest field ligand among the given options.

(B) According to the spectrochemical series, the order of field strength for the given ligands is: $S^{2-} < NCS^{-} < H_2O < NH_3$.
Among the given options, $NH_3$ has the highest field strength as it is a stronger field ligand compared to $H_2O$, $NCS^{-}$, and $S^{2-}$.

(C) According to the spectrochemical series, the order of increasing field strength for the given ligands is:
$I^- < Br^- < S^{2-} < SCN^- < Cl^- < N_3 < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < EDTA^{4-} < NH_3 < en < CN^- < CO$.
Comparing the given ligands: $C_2O_4^{2-} < H_2O < EDTA^{4-} < en$.
Thus, option $(C)$ is correct.

(A) $1$. The central metal atom is $Fe$ (atomic number $26$). The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
$2$. In $[Fe(CO)_5]$, $CO$ is a strong field ligand, which causes pairing of electrons in the $3d$ orbital.
$3$. The $Fe$ atom in $[Fe(CO)_5]$ is in the $0$ oxidation state.
$4$. Due to the strong field ligand $CO$, the $4s$ electrons pair up into the $3d$ orbitals, leaving one $4s$, three $4p$, and one $3d$ orbital vacant.
$5$. These five orbitals hybridize to form five $dsp^3$ hybrid orbitals, resulting in a trigonal bipyramidal geometry.

(B) $1$. The central metal ion is $Mn^{3+}$. The atomic number of $Mn$ is $25$, so the electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.
$2$. $CN^-$ is a strong field ligand, which causes pairing of electrons in the $3d$ orbitals.
$3$. In $Mn^{3+}$, the $3d^4$ configuration becomes $t_{2g}^4 e_g^0$ after pairing, leaving two $3d$ orbitals vacant.
$4$. These two vacant $3d$ orbitals, one $4s$ orbital, and three $4p$ orbitals hybridize to form $d^2sp^3$ hybrid orbitals.
$5$. Since the coordination number is $6$, the geometry is octahedral.

(C) $1$. The coordination complex is $K_4[Fe(CN)_6]$.
$2$. The central metal ion is $Fe^{2+}$.
$3$. In coordination complexes, the central metal ion accepts lone pairs of electrons from the ligands ($CN^-$ ions).
$4$. According to the Lewis theory, an electron pair acceptor is defined as a Lewis acid.
$5$. Therefore, $Fe^{2+}$ acts as a Lewis acid.

(D) $1$. The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$, so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$2$. $NH_3$ is a strong field ligand, which causes pairing of electrons in the $3d$ orbitals.
$3$. In the presence of $NH_3$, the six $3d$ electrons pair up in the first three $3d$ orbitals, leaving no unpaired electrons $(n = 0)$.
$4$. The hybridization is $d^2sp^3$, which corresponds to an octahedral geometry.