3. An element has its electronic configuration as 2, 8, 2. Now answer the following questions.

a. What is the atomic number of this element?

Answer. The atomic number of this element is 2 + 8 + 2 = 12.

b. What is the group of this element.

Answer. The electronic configuration of the given element is 2, 8. 2 which shows that this element has 2 valence electrons. Hence the given element is placed in group 2.

c. To which period does this element belong?

Answer. The electronic configuration of the given element is 2, 8, 2 which shows that the electrons are arranged in 3 shells. Hence the given element is placed in third period.

d. With which of the following elements would this element resemble?

(Atomic numbers are given in the brackets) N (7), Be (4) , Ar (18), Cl (17).  

Answer. Electronic Configuration:

N (7) – 2 , 5         Be (4) – 2, 2 Ar (18) – 2 , 8, 8    Cl (17) – 2, 8, 7

The element Beryllium (Be) has valence electron 2 just like the given element. Hence this element will be chemically similar to the given element which has the same number of valence electrons in its atoms.

If the electronic configuration of metal A is (2, 8, 1) and that of metal B is (2, 8, 2) then:

a. Which metal is less reactive?

Ans. Metal B is less reactive.

b. Write the names of the two metals.

Ans. Metal A is sodium & Metal B is magnesium.

c. Write the balanced chemical equation of reaction of any one metal with hydrochloric acid.

Ans. Mg(s) + 2HCl(aq) ----------> MgCl2(aq)  + H2(g)

LOGIC EX. NO. 1.5

1. If  B = {5, 6, 8, 10}, determine the truth value of each of the following.

i.               ∃x∈B, such that 3x+4=28.

ii.              ∀x∈B,x+7<14.

iii.             ∀x∈B,4x-3≥17.

iv.             ∃x∈B,such that x is even.

v.              ∃y∈B,such that (y-10)∈N.

2. Use quantifiers to convert each of the following open sentences defined on N, into true statement.

i.             x²=36

ii.             5x-3<10

iii.            x-7=9

iv.            y²+3≤7

v.             y²-11y+30=0

vi.            x²≥1

(ix) 2, 4, 8, 16, ..... (1 mark)

(ix) 2, 4, 8, 16, ..... (1 mark)

Sol. t₁ = 2¹ = 2

t₂ = 2² = 4

t₃ = 2³ = 8

t₄ = 2⁴ = 16

t₅ = 2⁵ = 32

t₆ = 2⁶ = 64

t₇ = 2⁷ = 128

t₈ = 2⁸ = 256

∴  The next four terms of the sequence are 32, 64, 128 and 256.

(vii) 2, 5, 8, 11, .... (1 mark)

(vii) 2, 5, 8, 11, .... (1 mark)

Sol. t₁ = 2

t₂ = 2 + 3 = 5

t₃ = 5 + 3 = 8

t₄ = 8 + 3 = 11

t₅ = 11 + 3 = 14

t₆ = 14 + 3 = 17

t₇ = 17 + 3 = 20

t₈ = 20 + 3 = 23

∴ The next four terms of the sequence are 14, 17, 20 and 23.

Metal A has electronic configuration of 2,8,1 and metal B has 2,8,8,2 which is more reactive. Identify these metals and vie their reactions with dil HCl.

Ans. Metal A is Sodium (Na) : 2,8,1

Metal B is Calcium (Ca) : 2,8,8,2.

Metal A (Sodium) is more reactive than metal B (Calcium).

Reaction: -

i.      Sodium metal reacts violently with dilute hydrochloric acid to form sodium chlorides and hydrogen.

2Na_(s) + 2HCl_(aq) → 2NaCl_(aq) + H_2(g)

ii.   Calcium reacts less vigorously to form calcium chloride and hydrogen.

Ca + 2HCl → CaCl₂ + H₂

Find the next two terms of the given sequence 2, 4, 8, 16, ...........

Find tn of the given A.P. : 3, 8, 13, 18, .......