Chemistry Classification of Elements and Periodicity in Properties

Anions have larger radii than atoms. Also, higher the e/p ratio higher the ionic radii. So, N? ³ > O? ² > F?

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)

Electronic configuration of Fe is [Ar] 4s²3d⁶ and in +3 oxidation state it has [Ar] 4s⁰3d⁵ configuration.

Al < Mg < Si < S < P

1^st I. E increase along period with exception on moving from group 2 to group 13 and group 15 to group 16

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

As $Q_c>K_c$ , so reaction has a tendency to move backward.

Except Te, all are metals.

Given electronic configuration is for Ga and in 5^th period diagonally situated element is Sn with respect to Ga. Hence electronic configuration of Sn is [Kr]4d¹⁰5s²5p².

Assertion (A) is correct & Reason (R) is incorrect

→character decreases from left to right & non metallic character increases from left to night

→I.E. increases & electron gain enthalpy also increases from left to right.

$A_3{B}_{2\left(s\right)}?3A_{\left(aq\right)}^{+2}+2B_{\left(aq\right)}^{3-}$

Solubility $\frac{x}{M}$ $\frac{3x}{M}$ $\frac{x}{M}$

$\therefore Ksp={\left[A^{+2}\right]}^3{\left[B^{-3}\right]}^2$

$={\left(\frac{3x}{M}\right)}^3{\left(\frac{2x}{M}\right)}^2$

$=108{\left(\frac{x}{M}\right)}^5$

$Ksp=a{\left(\frac{x}{M}\right)}^5=108{\left(\frac{x}{M}\right)}^5$

Ans. $\therefore$ a = 108