Chapter 2: Structure of Atom

Ans: According to de Broglie’s equation, $\lambda =\frac{h}{mv}$

Given, the K.E. of electrons is$3.0\times {{10}^{-25}}J$.

Since, $K.E=\frac{1}{2}m{{v}^{2}}$ 

Velocity (v) = $\sqrt{\frac{2K.E}{m}}$ 

\[\Rightarrow v=\sqrt{\frac{2(3.0\times {{10}^{-25}}J)}{9.10939\times {{10}^{-31}}kg}=}\sqrt{6.5866\times {{10}^{4}}}\] 

V= $811.579m{{s}^{-1}}$

 Substituting the value in the expression of $\lambda $ :

\[\lambda =\frac{6.626\times {{10}^{-34}}Js}{(9.10939\times {{10}^{-31}}kg)(811.579m{{s}^{-1}})}\] 

\[\lambda =8.9625\times {{10}^{-7}}m\] 

Ans: Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na) = 11

Number of electrons in ($N{{a}^{+}}$) = 10

A positive charge denotes the loss of an electron.

Similarly, 

Number of electrons in ${{K}^{+}}$  = 18 

Number of electrons in $M{{g}^{+2}}$  = 10 

Number of electrons in $C{{a}^{+2}}$  = 18 

A negative charge denotes the gain of an electron by a species. 

Number of electrons in sulphur (S) = 16 

$\therefore $ Number of electrons in ${{S}^{-2}}$  = 18 

Number of electrons in argon (Ar) = 18 

Hence, the following are isoelectronic species: 

(1) $N{{a}^{+}}$and $M{{g}^{+2}}$ (10 electrons each) 

(2) ${{K}^{+}}$,$C{{a}^{+2}}$, ${{S}^{-2}}$and Ar (18 electrons each).

Ans: The electronic configuration of H atom is $1{{s}^{1}}$ 

A negative charge on the species indicates the gain of an electron by it.

$\therefore $ Electronic configuration of ${{H}^{-}}$ is $1{{s}^{2}}$ 

Ans: The electronic configuration of Na atom is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$ 

A positive charge on the species indicates the loss of an electron by it.

$\therefore $Electronic configuration of $N{{a}^{+}}$is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{0}}or1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ 

Ans: The electronic configuration of O atom is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$ 

A negative charge on the species indicates the gain of two electrons by it.

$\therefore $Electronic configuration of ${{O}^{-2}}$is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

Ans: The electronic configuration of F atom is $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$ 

A negative charge on the species indicates the gain of an electron by it.

$\therefore $Electronic configuration of ${{F}^{-}}$is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

(ii). What are the Atomic Numbers of Elements Whose Outermost Electrons are Represented by 

Ans: Completing the electron configuration of the element as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$

$\therefore $ Number of electrons present in the atom of the element 

= 2 + 2 + 6 + 1 = 11

$\therefore $Atomic number of the element = 11

Ans: Completing the electron configuration of the element as $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$

$\therefore $ Number of electrons present in the atom of the element 

= 2+2+3=7

$\therefore $Atomic number of the element = 7

Ans: Completing the electron configuration of the element as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$

$\therefore $ Number of electrons present in the atom of the element 

= 2+2+6+2+5=17

$\therefore $Atomic number of the element = 17

(iii).Which Atoms are Indicated by the Following Configurations?

Ans: The electronic configuration of the element is $[He]2{{s}^{1}}$= $1{{s}^{2}}2{{s}^{1}}$ 

$\therefore $ Atomic number of the element = 3 

Hence, the element with the electronic configuration $[He]2{{s}^{1}}$is lithium (Li).

Ans:  The electronic configuration of the element is $[Ne]3{{s}^{2}}3{{p}^{3}}$= $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$ . 

$\therefore $ Atomic number of the element = 15 

Hence, the element with the electronic configuration $[Ne]3{{s}^{2}}3{{p}^{3}}$is phosphorus (P).

Ans: The electronic configuration of the element is $[Ar]4{{s}^{2}}3{{d}^{1}}$=$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{1}}$  . 

$\therefore $Atomic number of the element = 21 

Hence, the element with the electronic configuration $[Ar]4{{s}^{2}}3{{d}^{1}}$is scandium (Sc).

Ans: For g-orbitals, l = 4. 

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1). 

$\therefore $ For l = 4, minimum value of n = 5.

Ans: For the 3d orbital: 

Principal quantum number (n) = 3 

Azimuthal quantum number (l) = 2 

Magnetic quantum number (${{m}_{l}}$) = –2, –1, 0, 1, 2

Ans: For an atom to be neutral, the number of protons is equal to the number of electrons. 

$\therefore $ Number of protons in the atom of the given element = 29

Ans: The electronic configuration of the atom is $1{{s}^{2}}2{{s}^{2}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}$ 

The name of the element is Copper $_{29}^{35}Cu$ 

Ans: ${{H}_{2}}^{+}$:

Number of electrons present in hydrogen molecule (${{H}_{2}}$) = 1 + 1 = 2

$\therefore$Number of electrons in ${{H}_{2}}^{+}$= 2 – 1 = 1

${{H}_{2}}$:

Number of electrons in${{H}_{2}}$ = 1 + 1 = 2

${{O}_{2}}^{+}$:

Number of electrons present in oxygen molecule (${{O}_{2}}$) = 8 + 8 = 16 

$\therefore$ Number of electrons in ${{O}_{2}}^{+}$= 16-1 = 15

Ans: n = 3 (Given) 

For a given value of n, l can have values from 0 to (n – 1). 

$\therefore$For n = 3 

l = 0, 1, 2 

For a given value of l, ml can have (2l + 1) values. 

$\therefore$For l = 0, m = 0 

l = 1, m = –1, 0, 1 

l = 2, m = –2, –1, 0, 1, 2 

$\therefore$For n = 3 

l = 0, 1, 2 

${{m}_{0}}$ = 0 

${{m}_{1}}$  = –1, 0, 1 

${{m}_{2}}$  = –2, –1, 0, 1, 2

Ans: For 3d orbital, l = 2. 

For a given value of l, ${{m}_{l}}$ can have (2l + 1) values i.e., 5 values. 

 $\therefore$For l = 2 

${{m}_{2}}$ = –2, –1, 0, 1, 2

(iii). Which of the following orbitals are possible? 

1p, 2s, 2p and 3f

Ans: Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1. 

For a given value of n, l can have values from zero to (n – 1). 

Therefore for l is equal to 1, the minimum value of n is 2. 

Similarly, 

For f-orbital, l = 4. 

For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.

Ans: n = 1, l = 0 (Given) the orbital is 1s.

Ans: For n = 3 and l = 1 the orbital is 3p.

Ans: For n = 4 and l = 2 the orbital is 4d.

Ans: For n = 4 and l = 3 the orbital is 4f.

Ans:

a)  The given set of quantum numbers is not possible because the value of the   principal quantum number (n) cannot be zero.

b)  The given set of quantum numbers is possible.

c)  The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.

d)  The given set of quantum numbers is possible.

e)  The given set of quantum numbers is not possible. For n = 3, 

l = 0 to (3 – 1) 

l = 0 to 2 i.e., 0, 1, 2

f)  The given set of quantum numbers is possible.