The number of electrons in the species ${\mathrm{H}}_2^{+}$ $,$ ${\mathrm{H}}_2$ $,$ ${\mathrm{O}}_2^{+}$,  are respectively:

1. 15, 2, 1

2. 15, 1, 2

3. 1, 2, 15

4. 2, 1, 15

Orbital that does not exist:

The set of quantum numbers which represent 3p is
1. n =1, l =0;
2, n = 3; l=1
3. n = 4; l =2;
4. n = 4; l = 3 

Which of the following sets of quantum numbers is possible-
1. n = 0, l = 0, m_l = 0, m_s = + ½
2. n = 1, l = 0, m_l = 0, m_s = – ½
3. n = 1, l = 1, m_l = 0, m_s = + ½
4. n = 3, l = 3, m_l = –3, m_s = + ½

The total number of electrons in an atom with the following quantum numbers would be
(a) n = 4, m_s = – ½  (b) n = 3, l = 0
1. 16,  2
2. 11, 8
3. 16, 8
4. 12, 7

The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from n = 4 to n = 2 of He⁺ spectrum is -

1. ${\mathrm{n}}_1$ $=$ $3$ $\mathrm{to}$ ${\mathrm{n}}_2$ $=$ $4$

2. ${\mathrm{n}}_2$  = 3 to n₁ = 2

3. ${\mathrm{n}}_2$  = 3 to n₁ = 1

4. ${\mathrm{n}}_2$  = 2 to n₁ = 1

2 ×10⁸ atoms of carbon are arranged side by side. The radius of a carbon atom if the length of this arrangement is 2.4 cm would be 

1. 7.0 x 10^-11 m

2. 5.0 x 10^-11 m

3. 8.0 x 10^-11 m

4. 6.0 x 10^-11 m

The diameter of a zinc atom is 2.6 Å. If zinc atoms are arranged side by side lengthwise, number of atoms present in a length of 1.6 cm would be:

A certain particle carries 2.5 × 10^–16 C of static electric charge. The number of electrons present in it would be:

In Rutherford's experiment, generally, the thin foil of heavy atoms like gold, platinum, etc. have been used to be bombarded by the α-particles.

 If the thin foil of light atoms like aluminum etc. is used in Rutherford’s experiment, the difference that would be observed from the above results is :

1. The same results will be observed.

2. More deflection would be observed.

3. There will not be enough deflection.

4. None of the above.