$(i)$ $n\, = 4, l\, = 1$ $(ii)$ $n\, = 4, l\, = 0$
$(iii)$ $n\, = 3, l\, = 2$ $(iv)$ $n\, = 3, l\, = 1$
The sequence representing increasing order of energy, is
$(iii)$ $3\,d$ $(iv)$ $3\,p$
According to Bohr Bury's $(n+l)$ rule, increasing order of energy will be $(iv) < (ii) < (iii) < (i)$.
Note : If the two orbitals have same value of $(n+l)$ then the orbital with lower value of $n$ will be filled first.
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$(1)$ $C{H_3} - \mathop {CH}\limits^ \bullet C{H_3}$
$(2)$ (fig.)
$(3)$ $\mathop {C{H_2}}\limits^ \bullet - CH{(C{H_3})_2}$
$(4)$ $\mathop {C{H_2}}\limits^ \bullet - C{H_3}$
What is the major product i.e. $X$ in the reaction?
