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:

  1. (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) $4 p$



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