To estimate the effective nuclear charge (( Z_{\\text{eff}} )) for bromine (( \\text{Br} )) and bromide ion (( \\text{Br}^- )), we use Slater\u2019s rules to approximate the shielding effect and apply the formula:<\/p>\n\n\n\n
[
Z_{\\text{eff}} = Z – S
]<\/p>\n\n\n\n
where:<\/p>\n\n\n\n
- \n
- ( Z ) is the atomic number,<\/li>\n\n\n\n
- ( S ) is the shielding constant.<\/li>\n<\/ul>\n\n\n\n
Step 1: Determine Atomic Number ( Z )<\/strong><\/h3>\n\n\n\n
Bromine has ( Z = 35 ).<\/p>\n\n\n\n
Step 2: Determine Electron Configuration<\/strong><\/h3>\n\n\n\n
- \n
- Neutral Bromine (( \\text{Br} ))<\/strong>:
[
1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^5
]<\/li>\n\n\n\n- Bromide Ion (( \\text{Br}^- ))<\/strong>:
[
1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6
]<\/li>\n<\/ul>\n\n\n\nStep 3: Apply Slater\u2019s Rules to Find Shielding Constant ( S )<\/strong><\/h3>\n\n\n\n
Using Slater\u2019s rules, the shielding constant is estimated as follows:<\/p>\n\n\n\n
For Neutral Bromine (( \\text{Br} ))<\/strong><\/h4>\n\n\n\n
- \n
- ( 1s^2 ) contributes 0.85 \u00d7 2<\/strong> = 1.70<\/li>\n\n\n\n
- ( 2s^2, 2p^6 ) contribute 0.85 \u00d7 8<\/strong> = 6.80<\/li>\n\n\n\n
- ( 3s^2, 3p^6 ) contribute 0.85 \u00d7 8<\/strong> = 6.80<\/li>\n\n\n\n
- ( 3d^{10} ) contributes 1 \u00d7 10<\/strong> = 10.00<\/li>\n\n\n\n
- ( 4s^2, 4p^5 ) electrons shield each other with 0.35 \u00d7 6<\/strong> = 2.10<\/li>\n<\/ul>\n\n\n\n
Total shielding:
[
S_{\\text{Br}} = 1.70 + 6.80 + 6.80 + 10.00 + 2.10 = 27.40
]<\/p>\n\n\n\n[
Z_{\\text{eff}} (\\text{Br}) = 35 – 27.40 = 7.60
]<\/p>\n\n\n\nFor Bromide Ion (( \\text{Br}^- ))<\/strong><\/h4>\n\n\n\n
- \n
- Similar calculations, but with an additional 4p<\/strong> 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
]<\/li>\n<\/ul>\n\n\n\nFinal Answer<\/strong><\/h3>\n\n\n\n
[
Z_{\\text{eff}} (\\text{Br}), Z_{\\text{eff}} (\\text{Br}^-) = 7.60, 7.25
]<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
- \n
- Definition of Effective Nuclear Charge<\/strong>
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.<\/li>\n\n\n\n- Why Does ( Z_{\\text{eff}} ) Decrease for ( \\text{Br}^- )?<\/strong>
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.<\/li>\n\n\n\n- Impact on Atomic\/Ionic Radius<\/strong><\/li>\n<\/ol>\n\n\n\n
- \n
- Bromine Atom (120 pm)<\/strong>: Higher ( Z_{\\text{eff}} ) means stronger nuclear attraction, leading to a smaller atomic radius.<\/li>\n\n\n\n
- Bromide Ion (182 pm)<\/strong>: Lower ( Z_{\\text{eff}} ) means weaker nuclear attraction, allowing electrons to spread out, increasing the ionic radius.<\/li>\n<\/ul>\n\n\n\n
I will now generate an image representing the atomic and ionic radius relationship.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1654.png\")
<\/figure>\n\n\n\nHere 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!<\/p>\n","protected":false},"excerpt":{"rendered":"
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 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195494","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195494","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195494"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195494\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195494"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195494"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195494"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Bromide Ion (182 pm)<\/strong>: Lower ( Z_{\\text{eff}} ) means weaker nuclear attraction, allowing electrons to spread out, increasing the ionic radius.<\/li>\n<\/ul>\n\n\n\n
- Why Does ( Z_{\\text{eff}} ) Decrease for ( \\text{Br}^- )?<\/strong>
- Definition of Effective Nuclear Charge<\/strong>
- ( 2s^2, 2p^6 ) contribute 0.85 \u00d7 8<\/strong> = 6.80<\/li>\n\n\n\n
- Bromide Ion (( \\text{Br}^- ))<\/strong>:
- Neutral Bromine (( \\text{Br} ))<\/strong>: