Most stable cation of elements $113$ will be:

M^{3 +}

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M^{2 +}

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M^{+}

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M_{2}^{2 +}

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Solution

The correct option is A $M^{+}$
$113 : [R n] 4 f^{14} 5 d^{10} 7 s^{2} 7 p^{1}$
so it have 3 valence electron and it will show +1 and +3 oxidation state but due to inert pair effect +1 oxidation state wil be more stable.

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Similar questions

Consider the reaction . The standard reduction potential values of the elements and respectively. The order of their reducing power will be

[NCERT 1990]

M 1, M 2, M 3

M 4

$m^{3}$ $\geq$ $m^{2}$ . To satisfy this condition, $m$ can be

\frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}}

\frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}}

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