Question

The radial probability distribution curve obtained for an orbital wave function $\left(\Psi \right)$ has $3$ peaks and $2$ radial nodes. The valence electron of which one of the following metals does this wave function $\left(\Psi \right)$ correspond to?

• A. $Co$
• B. $Li$
• C. $K$
• D. $Na$

Answer 1

The correct option is D $Na$

Since the radial probability distribution curve has $3$ peaks, it means that for it $n=3$ and $l=0$. Moreover, it has two radial nodes ie, $n-l-1=2$, which also satisfied the previous condition, ie, $3-0-1=2$. Hence, the valence electron of the metal must present in $3s-orbital$

$Element$ $Configuration$

$C{u}_{29}$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^1$

$L{i}_3$ $1s^{2}2s^1$

$K_{19}$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

$N{a}_{11}$ $1s^{2}2s^{2}2p^{6}3s^1$

$\because$Only the valence electron of $N{a}_{11}$ is in $3s$ orbital.

$Hencethecorrectanswerisoption\left(A\right)$