MCQ
Element $"E"$ belongs to the period $4$ and group $16$ of the periodic table. The valence shell electron configuration of the element, which is just above $'E'$ in the group is $.....$
  • $3 s ^{2} \cdot 3 p ^{4}$
  • B
    $3 d ^{10} \cdot 4 s ^{2}, 4 p ^{4}$
  • C
    $4 d ^{10} \cdot 5 s ^{2}, 5 p ^{4}$
  • D
    $2 s ^{2}, p 4$

Answer

Correct option: A.
$3 s ^{2} \cdot 3 p ^{4}$
a
$E \Rightarrow[ Ar ] 3 d ^{10} 4 s ^{2} 4 p ^{4}$

Element above $E \Rightarrow[ Ne ] 3 s ^{2} 3 p ^{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

In the hydrocarbon,

$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2  \equiv \mathop C\limits_1 H$

The state of hybridization of carbons $1, 3$ and $5$ are in the following sequence

Quantum numbers of an atom can be defined on the basis of
Solubility product $(K_{sp})$ of salts of type $MX$, $MX_2$, $M_3X$ at temperature $T$ are $4 × 10^{-8}$, $3.2 × 10^{-14}$ and $2.7 × 10^{-15}$ respectively. Solubility of the salts at temperature $T$ are in the order
$\begin{matrix}
   ONa\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{3}}-C=C{{H}_{2}}\xrightarrow{HC\equiv CH}\xrightarrow{{{H}^{+}}}\xrightarrow[Pd-BaS{{O}_{4}}]{{{H}_{2}}}\xrightarrow[\Delta ]{A{{l}_{2}}{{O}_{3}}}  \\
\end{matrix}$

End product of the reaction is

If the electronic configuration of $M = 2,\,\,8,\,\,3$ and that of $A = 2,\,\,8,\,\,7,$ the formula of the compound is
Assign $A, B, C, D$ from given type of reaction. $Pb{O_2} + HCl\,\left( {dil} \right)\xrightarrow{{Warm}}PbC{l_2} \downarrow  + C{l_2} \uparrow  + {H_2}O$
If the density of a solution is $3.12\ gm/ml$, the mass of $1 5\ ml$ solution in significant figures is ............ $\mathrm{gm}$
When $400\, \mathrm{~mL}$ of $0.2\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$ solution is mixed with $600\, \mathrm{~mL}$ of $0.1\, \mathrm{M} \,\mathrm{NaOH}$ solution, the increase in temperature of the final solution is $....\,\times 10^{-2} \,\mathrm{~K}$. (Round off to the Nearest Integer).

$\left[\right.$ Use $: \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}: \Delta_{\mathrm{\gamma}} \mathrm{H}=-57.1\, \mathrm{k} \mathrm{J} \,\mathrm{mol}^{-1}$

Specific heat of $\mathrm{H}_{2} \mathrm{O}=4.18 \mathrm{Jk}^{-} \mathrm{g}^{-}$

density of $\mathrm{H}_{2} \mathrm{O}=1.0\, \mathrm{~g} \mathrm{~cm}^{-3}$

Assume no change in volume of solution on mixing.]

Which of the following are tautomers
The electrochemical extraction of aluminum from bauxite ore involves.

$(A)$ the reaction of $Al _2 O _3$ with coke ($C$) at a temperature $>2500^{\circ} C$.

$(B)$ the neutralization of aluminate solution by passing $CO _2$ gas to precipitate hydrated alumina $\left( Al _2 O _3 .3 H _2 O \right)$

$(C)$ the dissolution of $Al _2 O _3$ in hot aqueous $NaOH$.

$(D)$ the electrolysis of $Al _2 O _3$ mixed with $Na _3 AlF _6$ to give $Al$ and $CO _2$.