Question

Arrange in decreasing order, the energy of $2s$ orbital in the following atoms $H, Li, Na, K$

• A. $E_{2 s( H )}>E_{2 s( Li )}>E_{2 s( Na )}>E_{2 s( K )}$
• B. $E_{2 s( H )}>E_{2 s( Na )}>E_{2 s( Li )}>E_{2 s( K )}$
• C. $E_{2 s( H )}>E_{2 s( Na )}=E_{2 s( K )}>E_{2 s( Li )}$
• D. $E_{2 s( K )}< E_{2 s( Na )}< E_{2 s( Li )}< E_{2 s( H )}$

Answer 1

Answer: (A)

The electronic configuration of the given atoms is as
$H=1 s^{1} $
$L i=1 s^{2}, 2 s^{1} $
$N a=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{1}$
$K=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 4 s^{1}$
Thus, the increasing order of energy of $2 s$ orbitals is
$E_{2 s(K)}< E_{2 s(N a)}< E_{2 s(L i)}< E_{2 s(H)}$