Chemistry, asked by adithyapillai05, 8 months ago

Which of the following pairs has the highest bond order?
1) NO+ and CN+
2) CN- and NO+
3) CN- and CN+
4) O2^- and CN-​

Answers

Answered by Anonymous
9

Answer:

ᴏᴘᴛɪᴏɴ 32ɪs ᴛʜᴇ ʀɪɢʜᴛ ᴀɴsᴡᴇʀ

Explanation:

CN − - 14 ᴇʟᴇᴄᴛʀᴏɴs (6 ғʀᴏᴍ ᴄᴀʀʙᴏɴ, 7 ғʀᴏᴍ ɴɪᴛʀᴏɢᴇɴ, ᴀᴅᴅ 1 ғᴏʀ ɴᴇɢᴀᴛɪᴠᴇ ᴄʜᴀʀɢᴇ)

NO + - 14 ᴇʟᴇᴄᴛʀᴏɴs (8 ғʀᴏᴍ ᴏxʏɢᴇɴ, 7 ғʀᴏᴍ ɴɪᴛʀᴏɢᴇɴ, sᴜʙᴛʀᴀᴄᴛ 1 ғᴏʀ ᴘᴏsɪᴛɪᴠᴇ ᴄʜᴀʀɢᴇ)

CN + - 12 ᴇʟᴇᴄᴛʀᴏɴs (6 ғʀᴏᴍ ᴄᴀʀʙᴏɴ, 7 ғʀᴏᴍ ɴɪᴛʀᴏɢᴇɴ, sᴜʙᴛʀᴀᴄᴛ 1 ғᴏʀ ᴘᴏsɪᴛɪᴠᴇ ᴄʜᴀʀɢᴇ)

O 2− - 17 ᴇʟᴇᴄᴛʀᴏɴs (8 ғʀᴏᴍ ᴇᴀᴄʜ ᴏxʏɢᴇɴ, ᴀᴅᴅ 1 ғᴏʀ ɴᴇɢᴀᴛɪᴠᴇ ᴄʜᴀʀɢᴇ).

Sᴘᴇᴄɪᴇs ᴄᴏɴᴛᴀɪɴɪɴɢ ᴛʜᴇ sᴀᴍᴇ ɴᴜᴍʙᴇʀ ᴏғ ᴇʟᴇᴄᴛʀᴏɴs ʜᴀᴠᴇ ᴛʜᴇ sᴀᴍᴇ ʙᴏɴᴅ ᴏʀᴅᴇʀ.

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Answered by shadowsabers03
15

No. of electrons in \sf{NO^+=7+8-1=14.}

No. of electrons in \sf{CN^+=6+7-1=12.}

No. of electrons in \sf{CN^-=6+7+1=14.}

No. of electrons in \sf{O_2^-=2\times8+1=17.}

Molecular Orbital Configuration of compounds having 12 elements,

\sf{\longrightarrow M_{12}=\sigma1s^2<\sigma^*1s^2<\sigma2s^2<\sigma^*2s^2<\pi2p_x\,\!^2=\pi2p_y\,\!^2}

Here are 8 bonding electrons and 4 anti - bonding electrons.

  • \sf{n_b=8}
  • \sf{n_a=4}

Then bond order of compounds having 12 elements is,

\sf{\longrightarrow B_{12}=\dfrac{n_b-n_a}{2}}

\sf{\longrightarrow B_{12}=\dfrac{8-4}{2}}

\sf{\longrightarrow B_{12}=2}

Molecular Orbital Configuration of compounds having 14 elements,

\sf{\longrightarrow M_{14}=\sigma1s^2<\sigma^*1s^2<\sigma2s^2<\sigma^*2s^2<\pi2p_x\,\!^2=\pi2p_y\,\!^2<\sigma2p_z\,\!^2}

Here are 10 bonding electrons and 4 anti - bonding electrons.

  • \sf{n_b=10}
  • \sf{n_a=4}

Then bond order of compounds having 14 elements is,

\sf{\longrightarrow B_{14}=\dfrac{n_b-n_a}{2}}

\sf{\longrightarrow B_{14}=\dfrac{10-4}{2}}

\sf{\longrightarrow B_{14}=3}

Molecular Orbital Configuration of compounds having 17 elements,

\begin{aligned}\sf{\longrightarrow M_{17}}=\ \ &\sf{\sigma1s^2<\sigma^*1s^2<\sigma2s^2<\sigma^*2s^2<\sigma2p_z\,\!^2<\pi2p_x\,\!^2=\pi2p_y\,\!^2<}\\&\sf{\pi^*2p_x\,\!^2=\pi^*2p_y\,\!^1}\end{aligned}

Here are 10 bonding electrons and 7 anti - bonding electrons.

  • \sf{n_b=10}
  • \sf{n_a=7}

Then bond order of compounds having 17 elements is,

\sf{\longrightarrow B_{17}=\dfrac{n_b-n_a}{2}}

\sf{\longrightarrow B_{17}=\dfrac{10-7}{2}}

\sf{\longrightarrow B_{17}=1.5}

Decreasing order of bond order of these compounds is,

\sf{\longrightarrow B_{14}>B_{12}>B_{17}}

This implies \bf{CN^-} and \bf{NO^+} have the highest bond order.

Hence (2) is the answer.

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