MCQ
The maximum number of electrons in $p$-orbital with $n = 5,\,m = 1$ is
- A$6$
- ✓$2$
- C$14$
- D$10$
✓
Answer
Correct option: B.
$2$
b
(b) The maximum number of electron in any orbital is $ 2.$
(b) The maximum number of electron in any orbital is $ 2.$
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Similar questions
Match List $I$ with List $II$ :
→|
LIST $I$ (Compound / Species) |
LIST $II$ (Shape / Geometry) |
| $A$ $\mathrm{SF}_4$ | $I$ Tetrahedral |
| $B$ $\mathrm{BrF}_3$ | $II$ Pyramidal |
| $C$ $\mathrm{BrO}_3^{-}$ | $III$ See saw |
| $D$ $\mathrm{NH}_4^{+}$ | $IV$ Bent $T$-shape |
Choose the correct answer from the options given below :
Choose the correct product of this reaction
Which of the following is incorrect
Assertion : Cyclopentadienyl anion is much more stable than allyl anion.
Reason : Cyclopentadienyl anion is aromatic in character.
For a reaction ${H_{2\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \longrightarrow {H_2}{O_{\left( l \right)}}$ ; $\Delta {C_p} = 32\,J{K^{ - 1}}$, $\Delta H$ at $27\,^oC = -285.8\, kJ\, mol^{-1}$. What will be the value of $\Delta H$ at $127\,^oC$ ?.....$kJ/mol$
→$\begin{matrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
Ph-CH-C-H \\
|\,\,\,\,\,\,\,\,\,\, \\
OH\,\,\,\,\,\,\, \\
\end{matrix}\,\xrightarrow[{{H}_{2}}O]{H{{O}^{\Theta }}}\,(B)$ ; $(A)$ and $(B)$ are isomer and isomerization effectively carried out by trace of base $(B)$ . Identify $(B)$
→\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
Ph-CH-C-H \\
|\,\,\,\,\,\,\,\,\,\, \\
OH\,\,\,\,\,\,\, \\
\end{matrix}\,\xrightarrow[{{H}_{2}}O]{H{{O}^{\Theta }}}\,(B)$ ; $(A)$ and $(B)$ are isomer and isomerization effectively carried out by trace of base $(B)$ . Identify $(B)$
How many geometrical isomers are possible for the above compound ?
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→ethyne $\xrightarrow[\begin{smallmatrix}
(2)\,\,excess\,I-\,C{{H}_{2}}\,-\,{{(C{{H}_{2}})}_{2}}-\,C{{H}_{3}} \\
(3)\,\,{{H}^{\oplus }}
\end{smallmatrix}]{(1)\,\,excess\,NaN{{H}_{2}}}$
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→In reaction ${C_6}{H_5}C{H_3}\xrightarrow[{{\text{Oxidation}}}]{}A\xrightarrow[{NaOH}]{}B\xrightarrow{{{\text{Sodalime}}}}C$ Then $C$ is
→