Q14.163CP - Chemistry: Molecular Nature Of Matter And Change

Question

Account for the following facts: 

(a)   ${\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{Na}}^{\mathbf{+}}$ have very nearly the same radii.

(b) ${\mathbf{CaF}}_{\mathbf{2}}$ is insoluble in water, but NaF is quite soluble. 

(c) Molten ${\mathbf{BeCl}}_{\mathbf{2}}$ is a poor electrical conductor, whereas molten   is an excellent one

Step-by-Step Solution

Answer

1. ${\mathrm{Ca}}^{+2} \mathrm{and} {\mathrm{Na}}^{+}$having same radii because calcium loses two electrons and gain is an octet and similarly odium losses one electron to gain its octet and the now, the same almost same electronic configuration so having same radii 
2. Calcium fluoride is insoluble in water because fluoride is highly electronegative, and the bond is quite firm and can’t get a break, but the sodium fluoride bond is weak and breaks to form ions
3. Beryllium chloride form a covalent bond and calcium chloride forms an ionic bond, and an ionic bond is a good conductor of electricity.

Step 1: Ca + 2   and   Na + same radii

Atomic radii are defined as the size of an atom, and along the period, length decreases.

So, that calcium has small in size than sodium but ${\mathrm{Ca}}^{+2} \mathrm{and} {\mathrm{Na}}^{+}$ has the same size because calcium has an electronic configuration: data-custom-editor="chemistry" $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^2$and after losing two-electron to become stable and gain an octet, the electronic configuration becomes -  data-custom-editor="chemistry" $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^6$ and also in sodium electronic configuration is  data-custom-editor="chemistry" $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1$ and after losing one electron to gain octet it becomes - data-custom-editor="chemistry" $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^6$has the only difference is about the one shell. Still, they both are stable with complete octet hence having almost identical radii.

Step 2: CaF 2 is insoluble in water, but NaF is quite soluble.

Calcium fluoride is insoluble in water because fluoride is highly electronegative. The bond between calcium and fluoride is very strong in the crystal form, and the water molecules cannot break that strong bond. Hence, they are insoluble in the water here as sodium fluoride is soluble because sodium fluoride bonds are weak. In the water, the bond break, and the formation of sodium ions and fluoride ion occurs so that they are soluble in water. 

Step 3: Molten BeCl 2 is a poor electrical conductor, whereas molten CaCl is an excellent one

Beryllium chloride is a poor conductor of water because the bond formation is by the covalent bond that is by charging of electric, and they don’t form ions on breaking; hence, a poor conductor. In contrast, the ionic bond starts with calcium chloride and breaks from the ions, and the ionic is a good conductor of electricity; therefore, calcium chloride is a good conductor.