2 Marks Questions

Question 12 Marks

Write the electronic configuration of the following ions:
i. $H ^{-}$
ii. $Na ^{+}$
iii. $O ^{2-}$
iv. $F ^{-}$

Answer

i. $1 s^2$
ii. $1 s^2 2 s^2 2 p^6$
iii. $1 s^2 2 s^2 2 p ^6$
iv. $1 s^2 2 s^2 2 p^6$

Question 22 Marks

Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on the reduction of the 2-butyne show the geometrical isomerism?

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Question 32 Marks

Arrange the three isomers of pentane in increasing order of their boiling points.

Answer

2, 2-Dimethylpropane < 2-mehtylbutane < pentane.

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Question 42 Marks

How many seconds are there in 2 days?

Answer

Here, we know 1 day = 24 hours (h)
or $\frac{1 \text { day }}{24 h}=1=\frac{24 h}{1 \text { day }}$
then, $1 h=60 min$
or $\frac{1 h}{60 min}=1=\frac{60 min}{1 h}$
so, for converting 2 days to seconds,
i.e., 2 days _______ $=$ _______ seconds
The unit factors can be multiplied in series in one step only as follows:
2 day $\times \frac{24 h}{1 \text { day }} \times \frac{60 min}{1 h} \times \frac{60 s}{1 \min }$
$\begin{aligned} & =2 \times 24 \times 60 \times 60 s \\ & =172800 s\end{aligned}$

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Question 52 Marks

Which of the elements $Na , Mg , Si$ and P would have the greatest difference between the first and second ionisation enthalpies. Briefly explain your answer.

Answer

Among the given elements Na being alkali metal has only one electron in the valence shell, therefore has very low $\Delta_{ i } H _1$.
However, after the removal of one electron, it acquires nearest inert gas or neon gas configuration, i.e. $Na +\left(1 s^2, 2 s^2, 2 p ^2\right)$. Therefore, its $\Delta_{ i } H _2$ is expected to be very high. Consequently, the difference in first and second ionisation enthalpies would be greatest in case of Na.
However, it may be noted here that in case of $Mg , Si$ and P , although their $\Delta_{ i } H _1$ will be much higher than that of Na but their $\Delta_{ i } H _2$ be much lower than that of Na . As a result, the difference in their respectively, $\Delta_{ i } H _1$ and $\Delta_{ i } H _2$ would be much lower than that of Na .

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Question 62 Marks

A reaction between ammonia and boron trifluoride is given below:
$
: NH_3+B F_3 \rightarrow H_3 N: BF_3
$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridization of B and N in the reactants?

Answer

The acid is $BF _3$ and the base is $NH _3$. The Lewis theory of acids and bases explains it. The hybridization of B in $BF _3$ is $sp ^2$ and the hybridization of N in $NH _3$ is $sp ^3$.

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