The arrangement of orbitals on the basis of energy is based upon their $(n+l)$ value. Lower the value of $(n+l)$, lower is the energy. For orbitals having same values of $(n+l)$, the orbital with lower value of $n$ will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) $1 s, 2 s, 3 s, 2 p$
(b) $4 s, 3 s, 3 p, 4 d$
(c) $5 p, 4 d, 5 d, 4 f, 6 s$
(d) $5 f, 6 d, 7 s, 7 p$
II. Based upon the above information. Solve the questions given below.
(a) Which of the following orbitals has the lowest energy?
$$4 d, 4 f, 5 s, 5 p$$
(b) Which of the following orbitals has the highest energy?
$$5 p, 5 d, 5 f, 6 s, 6 p$$
I. (a) $(n+l)$ values of $1 s=1+0=1,2 s=2+0=2,3 s=3+0=3,2 p=2+1=3$
Hence, increasing order of their energy is
$$1 s<2 s<2 p<3 s$$
(b) $4 s=4+0=4,3 s=3+0=3,3 p=3+1=4, 4 d=4+2=6$
Hence, $3 s<3 p<4 s<4 d$
(c) $5 p=5+1=6,4 d=4+2=6,5 d=5+2=7,4 f=4+3=7,6 s=6+0=6$
Hence, $4 d<5 p<6 s<4 f<5 d$
(d) $5 f=5+3=8,6 d=6+2=8,7 s=7+0=7,7 p=7+1=8$.
Hence, $$7 s<5 f<6 d<7 p$$
II. (a) $(n+l)$ values of $4 d=4+2=6,4 f=4+3=7,5 s=5+0=5,7 p=7+1=8$
Hence, $5 s$ has the lowest energy.
$$\text { (b) } 5 p=5+1=6,5 d=5+2=7,5 f=5+3=8,6 s=6+0=6,6 p=6+1=7$$
Hence, $5f$ has highest energy.
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.
Neutron being neutral will not show deflection from the path on passing through an electric field.
Proton, cathode rays and electron being the charged particle will show deflection from the path on passing through an electric field.
An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?
An atom having atomic mass number 13 and number of neutrons 7 .
i.e., $$A=13, n=7$$
As we know that,
$$\begin{array}{ll} \text { As we know that, } & A=n+p \\ \therefore & p=A-n=13-7=6 \end{array}$$
Hence, $$Z=p=6$$
Wavelengths of different radiations are given below.
$\lambda(\mathrm{A})=300 \mathrm{~nm} \lambda$
$\lambda (\mathrm{B}) =300 \mu \mathrm{m} \lambda$
$\lambda (\mathrm{C})=3 \mathrm{~nm} \lambda$
$\lambda (\mathrm{D})=30 \mathop A\limits^o$
Arrange these radiations in the increasing order of their energies.
$$\begin{array}{ll} \lambda(A)=300 \mathrm{~nm}=300 \times 10^{-9} \mathrm{~m}, & \lambda(\mathrm{B})=300 \mu \mathrm{m}=300 \times 10^{-6} \mathrm{~m} \\ \lambda(\mathrm{C})=3 \mathrm{~nm}=3 \times 10^{-9} \mathrm{~m}, & \lambda(\mathrm{D})=30 \mathop A\limits^o=30 \times 10^{-10} \mathrm{~m}=3 \times 10^{-9} \mathrm{~m} \end{array}$$
$\begin{array}{ll}\text { Energy, } & E=\frac{h c}{\lambda} \\ \text { Therefore, } & E \propto \frac{1}{\lambda}\end{array}$
Increasing order of energy is $B< A< C=D$
The electronic configuration of valence shell of Cu is $3 d^{10} 4 s^1$ and not $3 d^9 4 s^2$. How is this configuration explained?
Configurations either exactly half-filled or fully filled orbitals are more stable due to symmetrical distribution of electrons and maximum exchange energy. In $3 d^{10} 4 s^1, d$-orbitals are completely filled and s-orbital is half-filled. Hence, it is more stable configuration.