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a
${\mathrm{Mg}}^{2+}<{\mathrm{Be}}^{2+}<{\mathrm{K}}^{+}<{\mathrm{Ca}}^{2+}$
b
${\mathrm{Be}}^{2+}<{\mathrm{K}}^{+}<{\mathrm{Ca}}^{2+}<{\mathrm{Mg}}^{2+}$
c
${\mathrm{K}}^{+}<{\mathrm{Ca}}^{2+}<{\mathrm{Mg}}^{2+}<{\mathrm{Be}}^{2+}$
d
${\mathrm{Ca}}^{2+}<{\mathrm{Mg}}^{2+}<{\mathrm{Be}}^{2+}<{\mathrm{K}}^{+}$

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detailed solution

Correct option is C

The charge/size ratio of a cation determines its polarizing power.
Higher is the charge and lower is the size, higher will be the charge to size ratios and higher will be the polarizing power of the cation.

The charge of K ion is +1 whereas that of other ions is +2. Thus, ${\mathrm{K}}^{+}$ has the lowest polarizing power.

For the remaining ions, the decreasing order of the size is

${\mathrm{Ca}}^{2+}>{\mathrm{Mg}}^{2+}>{\mathrm{Be}}^{2+}$

Hence, the increasing order of polarizing power is

${\mathrm{Ca}}^{2+}<{\mathrm{Mg}}^{2+}<{\mathrm{Be}}^{2+}$

Hence, the increasing order of the polarizing power of the cationic species is

${\mathrm{K}}^{+}<{\mathrm{Ca}}^{2+}<{\mathrm{Mg}}^{2+}<{\mathrm{Be}}^{2+}$.

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