Question

3. (6 points) Use principles of atomic structure and/or chemical bonding to answer each of the following. a. The radius of the Ca atom is 0.197 nm; the radius of the Ca ion is 0.099 nm. Account for this difference. b. The lattice energy of CaO(s) is -3,460 kJ/mol; the lattice energy for K2O(s) is -2,240 kJ/mol. Account for this difference.

Answer 1

Answer:-

(a)- As we know that the radius of Ca atom is 0.197 nm whereas radius of Ca²⁺ ion is 0.099 nm because size of cation (positive ion) of any atom is always smaller than its corresponding neutral atom. Since the no. of electrons in cation are less than that in neutral atom which results in increased nuclear charge (effective nuclear Charge i.e electrostatic force of attraction between nucleus proton and shells electrons) over the outermost electrons due to which size of cation is decreased. e.g.

Ca $\rightarrow$ Ca²⁺ + 2e-

20 p   20 p

20 e- 18 e-

20 n   20 n

Another reason for smaller size of Ca²⁺ ion as compared to the neutral Ca atom is that when removal of electrons from the neutral atom takes place then there is complete disappearance of the outermost shell i.e valence shell. Due to removal of outermost shell i.e valence shell size of cation (Ca²⁺) is also decreased.e.g.

Ca $\rightarrow$ Ca²⁺ + 2e-

No. of Shells :- 1 2 3 4

Ca (20) :- 2 , 8 , 8 , 2

atom

and

No. of Shells :- 1 2 3   

Ca²⁺ (18) :- 2 , 8 , 8 ,   

ion

As mentioned above in neutral Ca atom , there are 4 shells whereas Ca²⁺ ion has 3 shells.

(b)-

As we know that the lattice energy is the amount of energy released when oppositely charge ions get packed closely to form one mole of crystalline solid. Higher lattice energy means greater stability of ionic compound and greater attraction between cations (positive ion) and anions (negative ion).

Also we know that lattice energy depends upon the following factors.

1- If the size of ions is small, the inter-ionic attraction increases and the energy of the system decreases. This adds to the  strength and stability of ionic bond i.e increased lattice energy.

strength and stability of ionic bond $\alpha$ 1 / size of ions

2- Inter-ionic attraction increases with increase in charge on ions and hence increase lattice energy and vice versa.

strength and stability of ionic bond $\alpha$ charge on ions

therefore

As we know that size of K⁺ ion is 0.133 nm whereas size of Ca²⁺ ion is 0.099 nm .

So

size of Ca²⁺ ion < size of K⁺ ion

and

strength and stability of ionic bond of Ca²⁺ ion > strength and stability of ionic bond of K⁺ ion

Also we know that

Charge on Ca²⁺ ion is + 2 whereas charge on K⁺ ion is +1

So

Charge on Ca²⁺ ion > Charge on K⁺ ion

therefore

strength and stability of ionic bond of Ca²⁺ ion > strength and stability of ionic bond of K⁺ ion

that is why lattice energy of CaO_(s) is - 3460 kJ/mol which is higher than the lattice energy of K₂O_(s) having - 2240 kJ / mol because in CaO_(s) , Ca atom exists as Ca²⁺ ion whereas in K₂O_(s) , K atom exists as K⁺ ion.

CaO_(s)$\rightarrow$ Ca²⁺ + O^{- -}  

and

K₂O_(s)$\rightarrow$ 2K⁺ + O^{- -}