General Characteristics Questions in English

(D) copper coin is electroplated with zinc $(Zn)$ and then heated at high temperature.
During heating, zinc atoms diffuse into the copper lattice to form an alloy.
The resulting alloy is brass, which has a characteristic golden colour.
This occurs because the zinc migrates through the copper to form the $\alpha$-form of brass alloy (where the percentage of $Cu > 65 \%$ and $Zn < 35 \%$).
$Zn + Cu \xrightarrow{\Delta} \text{Brass (Golden colour)}$.
Thus, the correct option is $(D)$.

(D) The electronic configurations of cuprous $(Cu^{+})$ and cupric $(Cu^{2+})$ ions are as follows:
$Cu^{+} = [Ar] 3d^{10} 4s^{0}$
$Cu^{2+} = [Ar] 3d^{9} 4s^{0}$
Although $Cu^{+}$ has a more stable electronic configuration, $Cu^{2+}$ compounds are more stable in the solid state.
This is primarily because the $2^{nd}$ ionization potential $(IP)$ of $Cu$ is not sufficiently high to prevent the formation of $Cu^{2+}$, and the higher charge density of $Cu^{2+}$ leads to a significantly higher lattice energy in ionic compounds compared to $Cu^{+}$ compounds.
This large release of lattice energy compensates for the energy required to remove the second electron.

(D) Noble metals are defined by their resistance to oxidation and corrosion in moist air.
They are chemically inert towards many common reagents such as acids and bases.
This chemical stability is the primary reason they are classified as noble metals.

(A) Key Point: For elements with similar electronic configuration of the outermost shell, atomic size determines the value of ionisation energy. Larger size leads to lower ionisation energy.
Electronic configuration for $II^{nd}$ $I.E$:
$Zn^{+}$ ion $(Z=30) = [Ar] 3d^{10} 4s^{1} - (1734 \ kJ/mol)$
$Cd^{+}$ ion $(Z=48) = [Kr] 4d^{10} 5s^{1} - (1631 \ kJ/mol)$
$Hg^{+}$ ion $(Z=80) = [Xe] 4f^{14} 5d^{10} 6s^{1} - (1809 \ kJ/mol)$
Since the size of $Cd^{+} > Zn^{+}$, it has lower $II^{nd}$ ionisation energy. Due to the lanthanoid contraction (i.e.,poor screening by $4f$ and $5d$ electrons), $Hg^{+}$ has a higher $II^{nd}$ ionisation energy.
Hence, the correct order is $Zn > Cd < Hg$, and option $(A)$ is the correct answer.

(B) The highest oxidation state is achieved by $(n-1) d^5 n s^2$, which corresponds to a maximum oxidation state of $+7$. This is because the $(n-1) d$-electrons can participate in bonding along with $n s$-electrons, as their energy levels are comparable. The oxidation states for the given configurations are summarized below:
| Electronic configuration | Oxidation state |
| :--- | :--- |
| $(n-1) d^8 n s^2$ | $+2, +3, +4$ |
| $(n-1) d^3 n s^2$ | $+2, +3, +4, +5$ |
| $(n-1) d^5 n s^1$ | $+2, +3, +4, +5, +6$ |
| $(n-1) d^5 n s^2$ | $+2, +3, +4, +5, +6, +7$ |

(B) The electronic configurations are:
$Cr = [Ar] 3d^{5} 4s^{1}$
$Mn = [Ar] 3d^{5} 4s^{2}$
Statement-$I$: The first ionization enthalpy $(IE_{1})$ of $Cr$ is lower than that of $Mn$ because $Mn$ has a stable half-filled $d$-subshell and a full $s$-subshell, making it harder to remove an electron. Thus, Statement-$I$ is true.
Statement-$II$: For $IE_{2}$, $Cr$ loses its $4s^{1}$ electron to reach a stable $3d^{5}$ configuration, while $Mn$ loses its $4s^{1}$ electron from a $3d^{5} 4s^{2}$ configuration. However, $IE_{2}$ of $Cr$ is higher than $Mn$ because removing the second electron from $Cr$ disrupts the stable $d^{5}$ configuration. For $IE_{3}$, $Mn$ has a higher value because it involves removing an electron from the stable $d^{5}$ configuration of $Mn^{2+}$. Therefore, Statement-$II$ is false.

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$, where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{35} \text{ BM}$, we equate $\sqrt{n(n+2)} = \sqrt{35}$, which implies $n(n+2) = 35$.
Solving for $n$, we get $n^2 + 2n - 35 = 0$, which factors as $(n+7)(n-5) = 0$. Since $n$ must be positive, $n = 5$.
$A$ $Mn^{2+}$ ion has the electronic configuration $[Ar]3d^5$, which contains $5$ unpaired electrons.
Therefore, for $Mn^{2+}$, $\mu = \sqrt{5(5+2)} = \sqrt{35} \text{ BM}$.

(A) Magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$ $B$.$M$.,where $n$ is the number of unpaired electrons.
$1$. $\text{Cr}^{2+}: [Ar]3d^4 \implies n=4$, $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ $B$.$M$.
$2$. $\text{Ni}^{2+}: [Ar]3d^8 \implies n=2$, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ $B$.$M$.
$3$. $\text{Cu}^{2+}: [Ar]3d^9 \implies n=1$, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ $B$.$M$.
$4$. $\text{Co}^{2+}: [Ar]3d^7 \implies n=3$, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ $B$.$M$.
Comparing the values, $\text{Cr}^{2+}$ has the highest number of unpaired electrons $(n=4)$, and therefore, it has the highest magnetic moment.

(B) Interstitial compounds are formed when small atoms like $H, C, N$ are trapped inside the crystal lattice of transition metals.
Their key characteristics include:
$1$. They have high melting points, higher than those of pure metals.
$2$. They are very hard (some approach diamond in hardness).
$3$. They retain metallic conductivity.
$4$. They are chemically inert and usually non-stoichiometric.
Since they are extremely hard and possess metallic properties, the statement that they are 'very soft and ionic' is incorrect. Therefore, option $B$ is the correct answer.

(C) The highest manganese fluoride is $MnF_4$ (oxidation state of Mn is $+4$).
The highest manganese oxide is $Mn_2O_7$ (oxidation state of Mn is $+7$).
The difference $|x| = |7 - 4| = 3$.
We need to identify ions with $3$ unpaired electrons:
$Sc^{3+}$: $[Ar] 3d^0$, unpaired electrons = $0$.
$Zn^{2+}$: $[Ar] 3d^{10}$, unpaired electrons = $0$.
$V^{2+}$: $[Ar] 3d^3$, unpaired electrons = $3$.
$Fe^{2+}$: $[Ar] 3d^6$, unpaired electrons = $4$.
$Co^{2+}$: $[Ar] 3d^7$, unpaired electrons = $3$.
Thus, $V^{2+}$ and $Co^{2+}$ have $3$ unpaired electrons.

(D) The electronic configurations of the given elements are:
$Eu (63): [Xe] 4f^7 6s^2$
$Gd (64): [Xe] 4f^7 5d^1 6s^2$
$Dy (66): [Xe] 4f^{10} 6s^2$
$Ho (67): [Xe] 4f^{11} 6s^2$
$Yb (70): [Xe] 4f^{14} 6s^2$
$Lu (71): [Xe] 4f^{14} 5d^1 6s^2$
$Tm (69): [Xe] 4f^{13} 6s^2$
$Hf (72): [Xe] 4f^{14} 5d^2 6s^2$
Comparing the number of electrons in the $4f$ orbital:
Pair $A$ ($Eu$ and $Gd$): Both have $7$ electrons in the $4f$ orbital. (Correct)
Pair $B$ ($Dy$ and $Ho$): $Dy$ has $10$ and $Ho$ has $11$ electrons. (Incorrect)
Pair $C$ ($Yb$ and $Hf$): Both have $14$ electrons in the $4f$ orbital. (Correct)
Pair $D$ ($Lu$ and $Tm$): $Lu$ has $14$ and $Tm$ has $13$ electrons. (Incorrect)
Therefore, pairs $A$ and $C$ are correct.

(D) The borax bead test is used to identify transition metal ions that form coloured beads in the oxidising or reducing flame.
Among the given ions, $Fe^{2+}$, $Fe^{3+}$, and $Cr^{2+}$ are transition metal ions that produce coloured beads, whereas $Zn^{2+}$ forms a colourless bead and is considered to give a negative test.
Ionisation enthalpy generally increases with an increase in effective nuclear charge and stability of the electronic configuration.
$Fe^{3+}$ has a stable half-filled $d^5$ electronic configuration $([Ar] 3d^5)$ and a higher effective nuclear charge compared to $Fe^{2+}$ and $Cr^{2+}$, which results in the highest ionisation enthalpy among the choices provided.
Therefore, the correct ion is $Fe^{3+}$.

(A) An ion is paramagnetic if it has one or more unpaired electrons.
$1$. $Mn^{2+}$ $([Ar] 3d^5)$: $5$ unpaired electrons (Paramagnetic).
$2$. $Cu^{2+}$ $([Ar] 3d^9)$: $1$ unpaired electron (Paramagnetic).
$3$. $Zn^{2+}$ $([Ar] 3d^{10})$: $0$ unpaired electrons (Diamagnetic).
$4$. $Yb^{2+}$ $([Xe] 4f^{14})$: $0$ unpaired electrons (Diamagnetic).
$5$. $Sc^{3+}$ $([Ar] 3d^0)$: $0$ unpaired electrons (Diamagnetic).
$6$. $La^{3+}$ $([Xe] 4f^0)$: $0$ unpaired electrons (Diamagnetic).
$7$. $Gd^{3+}$ $([Xe] 4f^7)$: $7$ unpaired electrons (Paramagnetic).
$8$. $Lu^{3+}$ $([Xe] 4f^{14})$: $0$ unpaired electrons (Diamagnetic).
$9$. $Ti^{4+}$ $([Ar] 3d^0)$: $0$ unpaired electrons (Diamagnetic).
$10$. $Ce^{4+}$ $([Xe] 4f^0)$: $0$ unpaired electrons (Diamagnetic).
The paramagnetic ions are $Mn^{2+}$, $Cu^{2+}$, and $Gd^{3+}$.
Total number of paramagnetic ions = $3$.

(A) Transition metals exhibit high melting points due to the strong metallic bonding between their atoms.
This strength in metallic bonding arises because, in addition to the $ns$ electrons, the $(n-1)d$ electrons also participate in the formation of interatomic metallic bonds.
For $3d$ transition metals, both $4s$ and $3d$ electrons contribute to this bonding, leading to high enthalpy of atomization and consequently high melting points.
Therefore, both Assertion $A$ and Reason $R$ are correct, and Reason $R$ is the correct explanation for Assertion $A$.

(D) The $3d$ transition series corresponds to elements where the $3d$ orbital is being filled, which are elements with atomic numbers $21$ to $30$.
Electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
Electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
Both elements have an $ns^1$ (specifically $4s^1$) configuration and belong to the $3d$ transition series.
Therefore, the correct answer is $24$ and $29$.

(B) $1$. The atomic number of $Cu$ is $29$.
$2$. According to the Aufbau principle, the expected configuration is $[Ar] 3d^9 4s^2$.
$3$. However, a completely filled $d$-subshell $(d^{10})$ provides extra stability due to symmetry and exchange energy.
$4$. Therefore, one electron from the $4s$ orbital shifts to the $3d$ orbital, resulting in the observed configuration: $[Ar] 3d^{10} 4s^1$.

(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) 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 $C$.

(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 $(C)$.

(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.

(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 $(KMnO_4)$.

(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) $1$. 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) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
The electronic configuration of $Cr^{2+}$ ion is $[Ar] 3d^4 4s^0$.
Number of unpaired electrons $(n)$ = $4$.
The spin-only magnetic moment formula is $\mu = \sqrt{n(n+2)} \text{ BM}$.
Substituting $n = 4$ into the formula: $\mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \text{ BM}$.