Calculate the 'spin only' magnetic moment of $M ^{2+}(\operatorname{aq})$ ion $(Z=27)$
$Z=27$
$\Rightarrow[ Ar ] 3 d^{7} 4 s^{2}$
$\therefore M^{2+}=[A r] 3 d^{7}$
i.e., $3$ unpaired electrons
$\therefore n=3$
$\Rightarrow \sqrt{n(n+2)}=\mu$
$\Rightarrow \sqrt{3(3+2)}=\mu$
$\Rightarrow \sqrt{15}=\mu$
$\mu \approx 4 \,BM$
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is $25$
Identify the correct option
Many transition metals form interstitial compounds. The characteristics of these interstitial compounds are
$(I)$ They have high melting points, higher than those of pure metals
$(II)$ They are very hard
$(III)$ They retain metallic conductivity
$(IV)$ They are chemically more reactive than the pure metals
Hydrated $C{u^{ + 2}}$ ion will be
A transition metal ' $\mathrm{M}$ ' among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Fe}$ has the highest second ionisation enthalpy. The spin only magnetic moment value of $\mathrm{M}^{+}$ion is. . . . . . .$\mathrm{BM}$ (Near integer)
(Given atomic number $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}$ : $24, \mathrm{Mn}: 25, \mathrm{Fe}: 26)$