The atomic radius of a bromine atom is 120 pm, The ionic radius of the ion Br- is 182 pm.
Estimate effective nuclear charge Zeff for each of these species.
Express your answers numerically to three significant figures separated by a comma.
The Correct Answer and Explanation is :
To estimate the effective nuclear charge (( Z_{\text{eff}} )) for bromine (( \text{Br} )) and bromide ion (( \text{Br}^- )), we use Slater’s rules to approximate the shielding effect and apply the formula:
[
Z_{\text{eff}} = Z – S
]
where:
- ( Z ) is the atomic number,
- ( S ) is the shielding constant.
Step 1: Determine Atomic Number ( Z )
Bromine has ( Z = 35 ).
Step 2: Determine Electron Configuration
- Neutral Bromine (( \text{Br} )):
[
1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^5
] - Bromide Ion (( \text{Br}^- )):
[
1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6
]
Step 3: Apply Slater’s Rules to Find Shielding Constant ( S )
Using Slater’s rules, the shielding constant is estimated as follows:
For Neutral Bromine (( \text{Br} ))
- ( 1s^2 ) contributes 0.85 × 2 = 1.70
- ( 2s^2, 2p^6 ) contribute 0.85 × 8 = 6.80
- ( 3s^2, 3p^6 ) contribute 0.85 × 8 = 6.80
- ( 3d^{10} ) contributes 1 × 10 = 10.00
- ( 4s^2, 4p^5 ) electrons shield each other with 0.35 × 6 = 2.10
Total shielding:
[
S_{\text{Br}} = 1.70 + 6.80 + 6.80 + 10.00 + 2.10 = 27.40
]
[
Z_{\text{eff}} (\text{Br}) = 35 – 27.40 = 7.60
]
For Bromide Ion (( \text{Br}^- ))
- Similar calculations, but with an additional 4p electron:
[
S_{\text{Br}^-} = 1.70 + 6.80 + 6.80 + 10.00 + 2.45 = 27.75
]
[
Z_{\text{eff}} (\text{Br}^-) = 35 – 27.75 = 7.25
]
Final Answer
[
Z_{\text{eff}} (\text{Br}), Z_{\text{eff}} (\text{Br}^-) = 7.60, 7.25
]
Explanation
- Definition of Effective Nuclear Charge
The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. It accounts for electron shielding effects, which reduce the actual nuclear attraction felt by valence electrons. - Why Does ( Z_{\text{eff}} ) Decrease for ( \text{Br}^- )?
When bromine gains an extra electron to form ( \text{Br}^- ), electron-electron repulsion increases in the outermost shell. This slightly increases the shielding effect, reducing the effective nuclear charge. - Impact on Atomic/Ionic Radius
- Bromine Atom (120 pm): Higher ( Z_{\text{eff}} ) means stronger nuclear attraction, leading to a smaller atomic radius.
- Bromide Ion (182 pm): Lower ( Z_{\text{eff}} ) means weaker nuclear attraction, allowing electrons to spread out, increasing the ionic radius.
I will now generate an image representing the atomic and ionic radius relationship.
Here is the generated scientific illustration comparing the atomic radius of a bromine atom (120 pm) and the ionic radius of a bromide ion (182 pm). It visually represents the difference in size and the concept of effective nuclear charge (( Z_{\text{eff}} )). Let me know if you need any modifications or further explanations!