CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 7 - MCQExams.com

Atomic number of calcium is 20. Number of protons = number of electrons = 20

First shell can accommodate 2 electrons so $$K = 2$$.

Second shell can accommodate maximum of 8 electrons so $$L = 8$$.

Third shell can accommodate maximum of 18 electrons but if we put remaining 10 electrons it will violate the law that outermost electron can accommodate 8 electrons so it will accommodate only 8 electrons.

So number of electrons in M is 8. Now the remaining two will be in N.

So the configuration is $$2,8,8,2$$

The creation of NaOH is exothermic and will likely release enough energy to ignite the hydrogen gas, which will burn explosively with the oxygen in the atmosphere. Sodium reacts violently with water because it is much more active than hydrogen.

Here is the displacement reaction with water.

$$2Na\left( s \right) +2{ H }_{ 2 }O\left( l \right) \rightarrow 2Na^+{ (aq) }+2O{ H }^{ - }\left( aq \right) +{ H }_{ 2 }(g)$$

$$s=$$solid, $$l=$$liquid, $$aq=$$aqueous

Atomic number of sodium is 11. 

Number of protons = number of electrons = 11

First shell can accommodate 2 electrons so $$K = 2$$.

Second shell can accommodate maximum of 8 electrons so $$L = 8$$

Now remaining 1 electron will be accommodated in M i.e 1

So the configuration is $$2,8,1$$

Atomic number of chlorine is 17. Number of protons=number of electrons = 17

First shell can accommodate 2 electrons so $$K = 2$$.

Second shell can accommodate maximum of 8 electrons so $$L = 8$$

Third shell can accommodate 7 electrons. Therefore configuration is $$2,8,7$$

$$Max(M)=18, Max(N)=32, Max(L)=8, Max(K)=2$$.

Substituting the values of all these into the ratio we get,

$$\dfrac{(32+18)}{(2+8)}=5$$

$$Max(M)=18, Max(N)=32, Max(L)=8, Max(K)=2$$.

Now substituting the value in A we get

$$LHS =32+8=40$$

$$RHS =20\times 2=40$$

On moving from left to right in a period, the chemical reactivity of elements first decreases and then increases. On moving from left to right in a period, the number of valence electrons increases from 1 to 8. On moving from left to right in a period, the atomic size decreases; the size of the atoms decreases. Since the atomic size decrease when we move from left to right in a period, therefore, electronegativity increases in moving left to right. Atomic size increases when moving down in a group thus electronegativity decreases when we move from top to bottom in the periodic table. 

Metallic character decreases as you move across a period in the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell.

 Thus, electrons are filled in shells in the above manner. So,

For $$Be$$ : $$Z$$ = $$4$$ ; $$K$$ = $$2$$; $$L$$ = $$2$$

For $$O$$ : $$Z$$ = $$8$$ ; $$K$$ = $$2$$; $$L$$ = $$6$$

For $$S$$  : $$Z$$ :  $$16$$ ; $$K$$ = $$2$$; $$L$$ = $$8$$; $$M$$ = $$6$$

For $$P$$ : $$Z$$ = $$15$$ ; $$K$$ = $$2$$; $$L$$ = $$8$$; $$M$$ = $$5$$ 

The Potassium lone electron is farther away from the nucleus than the Sodium lon electron. So there is a less nuclear force to hold the Potassium electron, hence it can be ionized easier. 

The Potassium electron is in the fourth orbital, so there are three orbitals between it and the nucleus.

The outer electron in Potassium is in the $$4s$$ orbital, which is further away from the nucleus than the $$3s$$ orbital of the Sodium. This means that less energy is needed to remove the outermost electron and therefore the ionisation energy is lower.