Question
On the basis of data given below,
$$\eqalign{
& E_{\frac{{S{c^{3 + }}}}{{S{c^{2 + }}}}}^\Theta = - 0.37,E_{\frac{{M{n^{3 + }}}}{{M{n^{2 + }}}}}^\Theta = + 1.57 \cr
& E_{\frac{{C{r^{2 + }}}}{{Cr}}}^\Theta = - 0.90,E_{\frac{{C{u^{2 + }}}}{{Cu}}}^\Theta = 0.34 \cr} $$
Which of the following statements is incorrect ?
A.
$$S{c^{3 + }}$$ has good stability due of $$\left[ {Ar} \right]3{d^0}4{s^0}$$ configuration.
B.
$$M{n^{3 + }}$$ is more stable than $$M{n^{2 + }}.$$
C.
$$C{r^{2 + }}$$ is reducing in nature.
D.
Copper does not give $${H_2}$$ on reaction with $$dil.\,{H_2}S{O_4}.$$
Answer :
$$M{n^{3 + }}$$ is more stable than $$M{n^{2 + }}.$$
Solution :
$$M{n^{2 + }}\left( {{d^5}} \right)$$ is more stable than $$M{n^{3 + }}\left( {{d^4}} \right),$$ thus $$E_{\frac{{M{n^{3 + }}}}{{M{n^{2 + }}}}}^ - = + ve$$