The quantum numbers of four electrons are given below :
I. $$n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$$
II. $$n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$$
III. $$n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$$
IV. $$n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$$
The correct decreasing order of energy of these electrons is
The major product $$\mathrm{C}$$ in the below mentioned reaction is:
$$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[\Delta]{\text { alc. } \mathrm{KOH}} \mathrm{A} \xrightarrow{\mathrm{HBr}} \mathrm{B} \xrightarrow[\Delta]{\text { aq. } \mathrm{KOH}} \mathrm{C}$$
The compound that does not undergo Friedel-Crafts alkylation reaction but gives a positive carbylamine test is :
For an endothermic reaction:
(A) $$\mathrm{q}_{\mathrm{p}}$$ is negative.
(B) $$\Delta_{\mathrm{r}} \mathrm{H}$$ is positive.
(C) $$\Delta_r \mathrm{H}$$ is negative.
(D) $$\mathrm{q}_{\mathrm{p}}$$ is positive.
Choose the correct answer from the options given below: