Question

Write the electronic configuration of 24Cr and 29Cu.​

Answer 1

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$1. \: Cr (Z = 24) = 1s² \: 2s² \: 2p⁶ \: 3s² \: 4p⁶ \: 4s¹ \: 3d⁵$

$2. \: Cu (Z = 29) = 1s² \: 2s² \: 2p⁶ \: 3s² \: 4p⁶ \: 4s¹ \: 3d¹⁰$

ᴛʜᴇsᴇ ᴇʟᴇᴄᴛʀᴏɴɪᴄ ᴄᴏɴғɪɢᴜʀᴀᴛɪᴏɴs ᴀʀᴇ ᴜɴɪǫᴜᴇ sɪɴᴄᴇ ᴡɪᴛʜᴏᴜᴛ ғɪʟʟɪɴɢ ᴛʜᴇ s ᴏʀʙɪᴛᴀʟs ᴄᴏᴍᴘʟᴇᴛᴇʟʏ, ᴇʟᴇᴄᴛʀᴏɴs ᴇɴᴛᴇʀᴇᴅ ɪɴᴛᴏ -ᴅ ᴏʀʙɪᴛᴀʟs.

ʀᴇᴀsᴏɴ :-

ɪᴛ ʜᴀs ʙᴇᴇɴ sᴀɪᴅ ᴛʜᴀᴛ ᴡʜᴇɴᴇᴠᴇʀ ᴅ ᴏʀʙɪᴛᴀʟ ɪs ʜᴀʟғ ғɪʟʟᴇᴅ ᴏʀ ᴄᴏᴍᴘʟᴇᴛᴇʟʏ ғɪʟʟᴇᴅ ɪᴛ ɪs sᴛᴀʙʟᴇ. ᴛʜᴜs, ᴇʟᴇᴄᴛʀᴏɴs ʜɪᴛ ɴᴇᴀʀʟʏ ʜᴀʟғ ғɪʟʟᴇᴅ ᴀɴᴅ ᴄᴏᴍᴘʟᴇᴛᴇ ғɪʟʟᴇᴅ ɪɴ ᴛʜᴇ ɢɪᴠᴇɴ ɪɴsᴛᴀɴᴄᴇ, sᴏ ᴏɴᴇ ᴇʟᴇᴄᴛʀᴏɴ ᴀʀʀɪᴠᴇᴅ ғʀᴏᴍ ᴛʜᴇ s ᴏʀʙɪᴛᴀʟs ɪɴ ᴛʜᴇ ᴅ ᴏʀʙɪᴛᴀʟs ᴅᴜᴇ ᴛᴏ ᴛʜᴇ ᴀᴛᴛʀᴀᴄᴛɪᴏɴ.

Answer 2

Cu = 29 = 1s² 2s² 2p⁶ 3s² 4p⁶ 4s¹ 3d¹⁰ These electronic configuration are exceptional because electrons entered in 3-d orbitals without filling the 4s orbitals complete.