Atomic Structure - Result Question 49
####25. The electrons, identified by quantum numbers $n$ and $l$,
(i) $n=4, l=1$, (ii) $n=4, l=0$, (iii) $n=3, l=2$, (iv) $n=3, l=1$ can be placed in order of increasing energy, from the lowest to highest, as
(1999, 2M)
(a) (iv) $<$ (ii) $<$ (iii) $<$ (i)
(b) (ii) $<$ (iv) $<$ (i) $<$ (iii)
(c) (i) $<$ (iii) $<$ (ii) $<$ (iv)
(d) (iii) $<$ (i) $<$ (iv) $<$ (ii)
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Solution:
- (i) $n=4, l=1 \Rightarrow 4 p$-orbital
(ii) $n=4, l=0 \Rightarrow 4 s$-orbital
(iii) $n=3, l=2 \Rightarrow 3 d$-orbital
(iv) $n=3, l=1 \Rightarrow 3 d$-orbital
According to Aufbau principle, energies of above mentioned orbitals are in the order of
(iv) $3 p<$ (ii) $4 s<$ (iii) $3 d<$ (i) $4 p$
