Home
Class 12
CHEMISTRY
For a hypothetical case let value of l i...

For a hypothetical case let value of `l` is defined as `0,1,2,3,...(n+1)` for principle quantum number n

A

Atomic number of `1^(st)` noble gas is 8

B

Atomic number of `1^(st)` alkali metal is 9

C

Carbon has electron in `2p_z`

D

Element with atomic number 13 has half filled valance shell

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze the hypothetical case where the value of \( l \) is defined as \( 0, 1, 2, 3, \ldots, (n+1) \) for the principal quantum number \( n \). We will evaluate each option given in the question. ### Step-by-Step Solution: 1. **Understanding the Quantum Numbers**: - The principal quantum number \( n \) can take positive integer values (1, 2, 3, ...). - The azimuthal quantum number \( l \) can take values from \( 0 \) to \( n+1 \). Therefore, for \( n = 1 \), \( l \) can be \( 0, 1, 2 \). 2. **Assigning Orbitals**: - For \( l = 0 \): This corresponds to the \( s \) orbital. - For \( l = 1 \): This corresponds to the \( p \) orbital. - For \( l = 2 \): This corresponds to the \( d \) orbital. 3. **Calculating Electron Configuration**: - For \( n = 1 \), the hypothetical electron configuration would be: - \( 1s^2 \) (2 electrons) - \( 1p^6 \) (6 electrons) - \( 1d^{10} \) (10 electrons) - Total electrons = \( 2 + 6 + 10 = 18 \). 4. **Evaluating Each Option**: - **Option 1**: Atomic number of the first noble gas is 8. - The first noble gas is Helium (He) with an atomic number of 2, not 8. This option is incorrect. - **Option 2**: Atomic number of the first alkali metal is 9. - The first alkali metal is Lithium (Li) with an atomic number of 3. This option is incorrect. - **Option 3**: Carbon has electrons in 2pz. - Carbon (C) has an atomic number of 6, with the configuration \( 1s^2 2s^2 2p^2 \). It does not have electrons in \( 2p_z \) specifically, as it has \( 2p_x \) and \( 2p_y \) filled. This option is incorrect. - **Option 4**: Element with atomic number 13 has a half-filled valence shell. - The element with atomic number 13 is Aluminum (Al) with the configuration \( 1s^2 2s^2 2p^6 3s^2 3p^1 \). The \( 3p \) subshell has one electron, which is not half-filled. However, if we consider the hypothetical case, we can analyze the \( 1d^5 \) configuration, which is half-filled. This option is correct. 5. **Conclusion**: - The only correct option is **Option 4**: Element with atomic number 13 has half-filled valence shell.
Promotional Banner

Similar Questions

Explore conceptually related problems

Consider the hypothetical situation where the azimuthal quantum number , l,takes values 0, 1, 2, …….n+1 , where n is the principal quantum number . Then the element with atomic number :

An electron in a hydrogen atom makes a transition n_1 to n_2 where n_1 and n_2 are principle quantum numbers of the states . Assume the Bohr's model to be valid , the frequency of revolution in initial state is eight times that of final state. The ratio n n_1/n_2 is

The quantum number of the last electron of an element are below pralict the atomic number and following quantum number n = 3,l= 2, m = 0,s = -(1)/(2)

Assertion : The maximum number of electrons in the shell with principle quantum number n is equal to 2n^(2) . Reason : Two electrons can have the same value of three quantum numbers n, l and m_(l) , but must have the opposite spin quantum number.

Let sequence by defined by a_1=3,a_n=3a_(n-1)+1 for all n >1

The electron in a hydrogen atom makes a transition n_(1) rarr n_(2) , where n_(1) and n_(2) are the principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. the possible values of n_(1) and n_(2) are

The electron in a hydrogen atom makes a transition n_(1) rarr n_(2) , where n_(1) and n_(2) are the principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. the possible values of n_(1) and n_(2) are

Instead of principle quantum number (n), azimuthal quantum number (L) &magnetic quantum number m, a set of new quantum number s, t & u was introduce with similar logic but different value as defined below s=1,2,3,...........oo all positive integral values. t=(s^(2)-1^(1)),(s^(2)-2^(2)),(s^(2)-3^(3)) ......... No negative value u=((t+1))/2"to"+((t+1))/2 (including zero , if any ) in integral step . Each orbital can have maximum four electrons . (s+t) rule is defined ,similar to (n+l) rule. The number of subshells in which the third shell is sub divided equal to

For the principal quantum number n = 2, the possible values of azimuthal quantum number and magnetic quantum number respectively are

Find the maximum number of electrons are represented by quantum numbers n=3, l=1 .