MCQ
In covalency
- AElectrons are transferred
- ✓Electrons are equally shared
- CThe electron of one atom are shared between two atoms
- DNone of the above
✓
Answer
Correct option: B.
Electrons are equally shared
b
The covalency of an atom is the number of electron it shares with the other atoms when forming chemical bonds. Usually but not always, it is the number of covalent bonds it forms with the other bonds.
The covalency of an atom is the number of electron it shares with the other atoms when forming chemical bonds. Usually but not always, it is the number of covalent bonds it forms with the other bonds.
Example: In $CH _4$, carbon shares its four electrons with four electrons of four different $H$ - atoms to form $4$ covlent bonds. Hence covalency of carbon is $4.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A reaction $2A + B\xrightarrow{k}C + D$ is first order with respect to $A$ and $2^{nd}$ order with respect to $B$. Initial conc. $(t = 0)$ of $A$ is $C_0$ while $B$ is $2C_0$. If at $t = 30\,\min$ the conc. of $C$ is $C_0/4$ then rate expression at $t = 30\,\min$ is
→The number of precipitable halide ions in the sample $[Pt\left( {N{H_3}} \right)C{l_2}Br]Cl$ will be
→$A, B$ and $C$ are three biomolecules. The results of the tests performed on them are given below
→$\begin{array}{|c|c|c|c|}\hline {} &\text { Molisch's Test} & {\text { Barfoed Test}} & {\text { Biuret Test}} \\ \hline \text { A} & { Positive } & {\text { Negative }} & {\text { Negative }} \\ \hline \text { B } & {\text { Positive }} & {\text { Positive }} & {\text { Negative }} \\ \hline \text { C } & {\text { Negative }} & {\text { Negative }} & {\text { Positive }} \\ \hline\end{array}$
$A, B$ and $C$ are respectively
In the following reaction, final product is
A neutral atom will have the lowest ionization potential when its electronic configuration is
$\begin{array}{*{20}{c}}
{{H_3}C - CH - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $+HBr \rightarrow A$
→{{H_3}C - CH - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $+HBr \rightarrow A$
$A$ (predominantly) is
The addition of $HI$ in the presence of peroxide catalyst does not follow anti-Markovnikov's rule because
→For certain chemical reaction $X \rightarrow Y$, the rate of formation of product is plotted against the time as shown in the figure. The number of Correct statement/s from the following is $.......$.
→$(A)$ Over all order of this reaction is one
$(B)$ Order of this reaction can't be determined
$(C)$ In region$-I$ and $III$, the reaction is of first and zero order respectively
$(D)$ In region$-II$, the reaction is of first order
$(E)$ In region$-II$, the order of reaction is in the range of $0.1$ to $0.9$.
In the reaction,
→the product $(X)$ is
Which substance has the highest melting point?