- ASmall size of cation
- BLarge ionic Charge
- CVacant d orbitals
- DAll the above
Explanation:
Transition elements form complexes readily because
Due to lower size and higher charge, they have high charge density.
They have vacant d orbital in which a ligand can donate its electron and form complex.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $I$ | Column $II$ | ||
| $(A)$ | Kohlrausch law can calculate | $(P)$ | $\frac{{\Lambda _m^c}}{{\Lambda _m^o}}$ |
| $(B)$ | Molar conductance ${\Lambda _m}$ | $(Q)$ | $\frac{1}{R} \times \frac{l}{A}$ |
| $(C)$ | Specific conductance Kappa $\to (k)$ | $(R)$ | $\Lambda _m^o\,of\,c{a_3}{(P{O_4})_2}$ |
| $(D)$ | Degree of ionization of weak electrolyte | $(S)$ | $\frac{{k \times 1000}}{M}$ |
Which of the following option show correct matches
($1$) $W$ and $\mathbf{X}$ are, respectively
$[A]$ $O_3$ and $P_4 O_6$ $[B]$ $O_2$ and $P_4 O_6$ $[C]$ $O_2$ and $P_4 O_{10}$ $[D]$ $O_3$ and $P_4 O_{10}$
($2$) $Y$ and $Z$ are, respectively
$[A]$ $N_2 O_3$ and $\mathrm{H}_3 \mathrm{PO}_4$
$[B]$ $N_2 O_5$ and $\mathrm{HPO}_3$
$[C]$ $N_2 O_4$ and $HPO_3$
$[D]$ $N_2 O_4$ and $H_3 PO_3$
Give the answer of quetion ($1$) and ($2$)
