Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$$P{F_5}\,$$  trigonal bipyramidal

$$Br{F_5}$$  square pyramidal (distorted)

Total number of electrons in $$O_2^{2 - } = 16 + 2 = 18$$
According to $$MOT,$$  the configuration of $$O_2^{2 - }$$  is
$$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$    $$\sigma 2p_x^2,\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^2 \approx \mathop \pi \limits^ * 2P_z^2$$
So, the number of antibonding electron pairs = 4

$$\because $$ It forms hydrogen bonds with water

Isoelectronic species contain same number of electrons. $$B{e^{2 + }}$$  contains 2 electrons. Among the given options, only $$L{i^ + }$$ contains 2 electrons and therefore, it is isoelectronic with $$B{e^{2 + }}.$$
$$\eqalign{ & {H^ + } \to {\text{no}}\,{\text{electron}},\,N{a^ + } \to 10{e^ - } \cr & L{i^ + } \to 2{e^ - } \cr & M{g^{2 + }} \to 10{e^ - } \cr} $$
Hence, $$B{e^{2 + }}$$  is isoelectronic with $$L{i^ + }.$$

Presence of unpaired electrons makes a molecule paramagnetic. More the number of unpaired electrons, higher is the paramagnetism.

In covalent bonds between two identical non-metal atoms share the pair of electrons equally between them,
e.g. : $${F_2},{O_2},{N_2},$$

In case of single bond, there is only one $$\sigma - {\text{bond}},$$   in case of double bond, there is one $$\sigma $$  and one $$\pi - {\text{bonds}}$$   while in case of triple bond, there is one $$\sigma $$  and two $$\pi - {\text{bonds}}{\text{.}}$$   Thus, angular shape of ozone $$\left( {{O_3}} \right)$$  contains $${\text{2}}\sigma $$  and $$1\pi - {\text{bonds}}$$   as shown below

Since $$F$$ form $$H$$ - bond $${\left[ {H{F_2}} \right]^ - }$$  exists. Therefore $$KH{F_2}$$  gives $${K^ + } + HF_2^ - $$

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