Structure of Atom Test - 15

Electronic configuration of bromine is

\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^5\)

So, Total electrons in p orbital is

6 + 6 + 5 = 17

When the azimuthal quantum number has the value 2, the number of Solution :

Each sub shell of quantum number l contains 2l + 1 orbitals. Thus, if l = 2, then there are (2 × 2) + 1 = 5 orbitals

As for l shell n = 2 and for balmer series

n = 2 so it must be balmer series.

According to Hund's rule, orbitals of the same energy are each filled with one electron before filling any with a second.

Also, these first electrons have the same spin.

For fifth energy level, n = 5

So l = 0 to (n - 1) = 0, 1, 2, 3, 4

It will have 5s, 5p, 5d, 5f, 5g.

For s orbital, it has 1 orbital

For p orbital, it has 3 orbital

For d orbital, it has 5 orbital

For f orbital, it has 7 orbital

For g orbital, it has 9 orbital

Total no. of orbitals will be

= 1+ 3 + 5 + 7 + 9

= 25 orbitals.

Dually ionised Silicon (\(S^{-2}\))

No of protons = Atomic number (Silicon = 16)

No of electrons = Atomic number (16) + 2 electrons (anion) = 18

No of neutrons = Mass number - Atomic Number

= 32 - 16 = 16

There are \(n^2\) orbitals for each energy level. For n = 1, there is 12 or one orbital.

For n = 2, there are 22 or four orbitals.

For n = 3 there are nine orbitals, for n = 4 there are 16 orbitals

\(H^+\) (proton) will have very large hydration energy due to its very small ionic size.

Hydration energy \(\propto\) \(\frac{1}{Size}\).

A molecule of carbon dioxide contains one atom of carbon and two atoms of oxygen. Each carbon atom contains 6 electrons whereas each oxygen atom contains 8 electrons (equal to the atomic number of the element). A molecule of \(CO_2\) contains 22 electrons. 

According to de-Broglie equation

\(\lambda = \frac{h}{mv},\) p = mv,

\(\lambda =\frac{h}{p}\), \(\lambda =\frac{h}{mc}\)

\(\frac{1}{\lambda} = RZ^2\Big[\frac{1}{n_1^2} - \frac{1}{n_2^2}\Big]\)

R x \(2^2 \Big[\frac{1}{1^2} - \frac{1}{2^2} \Big]\)

⇒ 3R; \(\lambda = \frac{1}{3R}\)

\(Mn^{3+} [Ar] 3d^4\)

The number of radial nodes of 3s and 2p orbitals respectively are 2 and 0 respectively.

Fe (Z = 26): \(1s^22s^22p^63s^23p^64s^23d^2\)

\(Fe^{2+}\) (ferrous ion): \(1s^22s^22p^63s^23p^63d^6\)

Number of unpaired electrons = 4

(i) A plane passing through the nucleus, on which the probability of finding electron is zero, is called a nodal plane.

(ii) The number of nodal planes in an orbital is equal to azimuthal quantum number (l).

(iii) For d orbitals, azimuthal quantum number is 2. Therefore, d orbital has two nodal planes.

(i) The near UV region lies closest to visible light, and includes wavelengths between 200 and 400 nm.

(ii) The higher energy, shorter wavelength far UV region spans wavelengths between 91 and 200 nm.

Carbon,

Atomic Number = 6 = Electrons = Protons

Neutrons = Mass Number − Protons = 12 - 6 = 6

Silicon,

Atomic Number = 14 = Electrons = Protons

Neutrons = Mass Number - Protons = 28 - 14 = 14

Ratio = 6 : 14 or 3 : 7