
For each value of ${\text{l}}$, the possible values of ${m_l}$ are …………
(a) $\left( {2l + 1} \right)$
(b) ${\text{2l}}$
(c) $\left( {{\text{2l - 1}}} \right)$
(d) ${\text{2}}\left( {{\text{l + 1}}} \right)$
Answer
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Hint: ${\text{l}}$ is Azimuthal quantum number, and ${m_l}$ is Magnetic quantum number. The value of ${m_l}$ depends on the ${\text{l}}$ as for a given value of ${\text{l}}$, ${m_l}$ can have values from $ - l{\text{ to }} + l$, including zero.
Complete step by step answer:
The azimuthal quantum number, ${\text{l}}$ determines the orbital angular momentum of electrons, and also explains about the orbital shape. For given value of principal quantum number, ${\text{n}}$, ${\text{l}}$ can have values from $0{\text{ to }}\left( {n - 1} \right)$. The magnetic quantum number, ${m_l}$, tells about the number of orbitals present in a subshell, and their orientation. For a certain value of ${\text{l}}$, ${m_l}$ can have integer values from $ - l{\text{ to }} + l$, including $0$. This means that, for $l = 1$, ${m_l}$ will have values $ - 1,0, + 1$, and for $l = 2$, ${m_l}$ will be $ - 2, - 1,0, + 1, + 2$. Observing this trend, it can be obtained that for a given ${\text{l}}$, ${m_l}$ can have $\left( {2l + 1} \right)$ values. Now, it can be verified that for ${\text{s}}$ subshell, the value for ${\text{l}}$ is zero, which gives ${m_l} = \left( {2 \times 0 + 1} \right) = 1$ , and, so, $s$ has only one orbital. At $l = 1$, ${m_l} = \left( {2 \times 1 + 1} \right) = 3$, which is calculated above equals $ - 1,0, + 1$, and it goes on the same.
So, the correct answer is Option A .
Additional Information:
The first quantum number is called the principal number, ${\text{n}}$, it gives the shell number in which the electron is present, the second quantum number is called the azimuthal quantum number, ${\text{l}}$, it gives the shape of that particular shell in which electron is present, and the third quantum number is called the magnetic quantum number, it gives the way that orbital is held in the space.
Note:
The value of ${m_l}$ will decide the number of individual orbitals in a subshell. For $l = 1$; ${m_l} = - 1,0, + 1$. So, there are three p orbitals, namely ${p_{x,}}{p_y}{\text{ and }}{p_z}$. For $l = 2$ ; ${m_l} = - 2, - 1,0, + 1, + 2$, therefore, there are five d orbitals.
Complete step by step answer:
The azimuthal quantum number, ${\text{l}}$ determines the orbital angular momentum of electrons, and also explains about the orbital shape. For given value of principal quantum number, ${\text{n}}$, ${\text{l}}$ can have values from $0{\text{ to }}\left( {n - 1} \right)$. The magnetic quantum number, ${m_l}$, tells about the number of orbitals present in a subshell, and their orientation. For a certain value of ${\text{l}}$, ${m_l}$ can have integer values from $ - l{\text{ to }} + l$, including $0$. This means that, for $l = 1$, ${m_l}$ will have values $ - 1,0, + 1$, and for $l = 2$, ${m_l}$ will be $ - 2, - 1,0, + 1, + 2$. Observing this trend, it can be obtained that for a given ${\text{l}}$, ${m_l}$ can have $\left( {2l + 1} \right)$ values. Now, it can be verified that for ${\text{s}}$ subshell, the value for ${\text{l}}$ is zero, which gives ${m_l} = \left( {2 \times 0 + 1} \right) = 1$ , and, so, $s$ has only one orbital. At $l = 1$, ${m_l} = \left( {2 \times 1 + 1} \right) = 3$, which is calculated above equals $ - 1,0, + 1$, and it goes on the same.
So, the correct answer is Option A .
Additional Information:
The first quantum number is called the principal number, ${\text{n}}$, it gives the shell number in which the electron is present, the second quantum number is called the azimuthal quantum number, ${\text{l}}$, it gives the shape of that particular shell in which electron is present, and the third quantum number is called the magnetic quantum number, it gives the way that orbital is held in the space.
Note:
The value of ${m_l}$ will decide the number of individual orbitals in a subshell. For $l = 1$; ${m_l} = - 1,0, + 1$. So, there are three p orbitals, namely ${p_{x,}}{p_y}{\text{ and }}{p_z}$. For $l = 2$ ; ${m_l} = - 2, - 1,0, + 1, + 2$, therefore, there are five d orbitals.
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