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 × 10^–11 m

2. 5.0 × 10^–11 m

3. 8.0 × 10^–11 m

4. 6.0 × 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:
1. 5.153 × 10⁷
2. 6.153 × 10⁷
3. 4.153 × 10⁹
4. 6.153 × 10³

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 \(\alpha\)-particle scattering experiment, a thin foil of heavy metals like gold or platinum is typically used. If a lighter metal foil, such as aluminum, were used instead, how would the observed deflection patterns change?

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