NEET Practice Questions (MCQs) with Answers & Solutions

(c) R is p-block element, because difference between $I{E}_{2 }and I{E}_3$ is not very high as compared to between $I{E}_1 and I{E}_2; hence stable oxidation state of R will be higher than +2.$ 

(a) $$\begin{gathered} I.E. + E.A. =275+86=361 kcal mo{l}^{-1} \\ =361\times 4.184=1510.42 kJ mo{l}^{-1} \\ \therefore Electronegativity=\frac{1510.42}{540}=2.797=2.8 \end{gathered}$$

(d) 

   $$\begin{gathered} \left(a\right) \underset{4p^4}{Se}\stackrel{I.E{.}_1}{\to }\underset{4p^3}{S{e}^{+}}\stackrel{I.E{.}_2}{\to }\underset{4p^2}{S{e}^{2+}} \underset{4p^3}{As}\stackrel{I.E{.}_1}{\to }\underset{4p^2}{A{s}^{+}}\stackrel{I.E{.}_2}{\to }\underset{4p^1}{A{s}^{2+}} \\ \left(b\right) \underset{2p^2}{C}\to \underset{2p^1}{C^{+}}\to {\underset{2s^2}{C}}^{2+} \underset{2p^3}{N}\to {\underset{2p^2}{N}}^{+}\to {\underset{2p^1}{N}}^{2+} \underset{2p^4}{O}\to {\underset{2p^3}{O}}^{+}\to {\underset{2p^2}{O}}^{2+} \\ \left(c\right) \underset{2p^5}{F}\stackrel{I.E{.}_1}{\to }\underset{2p^4}{F^{+}}\stackrel{I.E{.}_2}{\to }{\underset{2p^3}{F}}^{2+}\stackrel{I.E{.}_3}{\to } F^{3+} \underset{2p^4}{O}\stackrel{I.E{.}_1}{\to }\underset{2p^3}{O^{+}}\stackrel{I.E{.}_2}{\to }{\underset{2p^2}{O}}^{2+}\stackrel{I.E{.}_3}{\to }{O}^{3+} \\ \left(d\right) In respective period, noble gases have higher than +2. \end{gathered}$$

(d) $$\begin{gathered} A \Rightarrow H\left(1s^1\right) \\ B \Rightarrow He\left(1s^2\right) \\ C \Rightarrow Li\left(1s^{2}2s^1\right) \\ {A}_1=I{E}_{1 } B_2=I{E}_2\left(B\right) \\ {B}_1=I{E}_1\left(B\right) C_2=I{E}_2\left(C\right) \\ {C}_1=I{E}_1\left(C\right) C_3=I{E}_3\left(C\right) \\ {B}_1>A_1>C_1 C_3>B_2>A_1 C_3>C_2>B_2 \\ He>H>Li L{i}^{2+} H{e}^{+} H L{i}^{2+ }L{i}^{+} H{e}^{+} \\ 1s^2 1s^1 2s^1 1s^1 1s^1 1s^1 1s^2 1s^2 1s^1 \end{gathered}$$

B

$$\begin{gathered} \left(a\right) \left[Ar\right]3d^{5}4s^1 \to Cr\left(24\right) \to 4^{th} period, 6^{th} group \\ \left(b\right) \left[Kr\right]4d^{10} \to Pd\left(46\right) \to 5^{th} period, {10}^{th} group \\ \left(c\right) \left[Rn\right]6d^{2}7s^2 \to Th\left(90\right) \to 7^{th} period, 3rd group \\ \left(d\right) \left[Xe\right]4f^{14}5d^{2}6s^2 \to Hf\left(72\right) \to 6^{th} period, 3^{rd} group \end{gathered}$$

(b)  Removal of two electrons (one by one ) from an atom requries energy = $I{P}_1$ + $I{P}_2$.

(c) Electron affinity order for halogen is $Cl>F>Br>I$.

(d) $E{A}_1$ for elements is exothermic and $E{A}_2$ is endothermic. Also $E{A}_2$ for $O>E{A}_1$ for O.

(D) (1) $M{g}^{2+}, N{a}^{+} and F^{-}$ are isoelectronic and thus follows the order ${}_{12}M{g}^{2+}<{}_{11}N{a}^{+}<{}_{9}F_{-}$.

Al belongs to third period and has no charge so it is largest.

$N{a}^{+}=102 pm; M{g}^{2+}=72 pm; Al =143 pm, F^{-}=133 pm.$

 $K^{+}$= has more number of shells than $M{g}^{2+} and A{l}^{3+} and M{g}^{2+}$ are isoelectronic but $A{l}^{3+}$ has higher nuclear charge so $A{l}^{3+}<M{g}^{2+}. M{g}^{2+} and L{i}^{+}$ has diagonal relationship. But due to +2 charge in $M{g}^{2+}, the M{g}^{2+}$ is smaller than $L{i}^{+}$. Hence $A{l}^{3+}$ is the smallest one.

$K^{+}=1.38 A, L{i}^{2+}=0.76 A, M{g}^{2+}=0.72 A and A{l}^{3+}=0.535 A.$

As the number of electrons are lost, the attraction between valence shell electrons and nucleus increases. As a consequence of this the electrons are pulled closer to the nucleus leading to the contraction in size of ions.

Across the period the nuclear charge increases and thus the size of atoms decreases.

Mg = 160 pm; Al = 143 pm; Si = 118; P = 110 pm.

(D) Due to small size of F atom, the electron-electron repulsions in compact 2-p sub shell are large and hence the incoming electron is not accepted with the same ease as is the case with Cl (less electron - electron repulsions)