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 = $3$.