MCQ
The quantum number/s needed to describe an electron fully in an atom is/are:
- A$1$
- B$2$
- C$3$
- ✓$4$
Thus all the four quantum numbers $n , l , m$ and $n$ are required. This is in accordance with Pauli's exclusion principle according to which, no two electrons in an atom can have the same set of four quantum numbers.
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| List $I$ (Substances) | List $II$ (Processes |
| $(A)$ Sulphuric acid | $(i)$ Haber's process |
| $(B)$ Steel | $(ii)$ Bessemer's process |
| $(C)$ Sodium hydroxide | $(iii)$ Leblanc process |
| $(D)$ Ammonia | $(iv)$ Contact process |
$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[{{\text{H}}_{2}}{{\text{O}}_{2}},\bar{O}\text{H}]{\text{B}{{\text{H}}_{3}},\text{THF}}(P)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(Q)$
$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[NaB{{H}_{4}}\,,\,H{{O}^{\Theta }}]{Hg\,{{(OAc)}_{2}}\,,\,{{H}_{2}}O}(R)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(S)$
$C{H_3} - CH(C{H_3}) - C{(C{H_3})_2} - C{H_2} - CH(C{H_3}) - C{H_2} - C{H_3}$
|
|
Primary |
Secondary |
Tertiary |
Quaternary |
|
$(a)$ |
$6$ |
$2$ |
$2$ |
$1$ |
|
$(b)$ |
$2$ |
$6$ |
$3$ |
$0$ |
|
$(c)$ |
$2$ |
$4$ |
$3$ |
$2$ |
|
$(d)$ |
$2$ |
$2$ |
$4$ |
$3$ |
$Z\,\xrightarrow{{PC{l_5}}}\,X\,\xrightarrow[\Delta ]{{{\text{Alc}}{\text{.KOH}}}}\,Y\,\xrightarrow{{{H_2}O/{H^ \oplus }}}\,\mathop Z\limits_{{\text{(Major)}}} \,,\,\,Z$ is