To solve the problem, we need to identify the element with the electronic configuration [Ar]3d\(^{10}\)4s\(^1\).
1. Understanding the Configuration:
The configuration [Ar] represents the electron configuration of argon, which is 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^6\), accounting for 18 electrons. The additional 3d\(^{10}\)4s\(^1\) indicates:
- 3d\(^{10}\): 10 electrons in the 3d subshell.
- 4s\(^1\): 1 electron in the 4s subshell.
Total electrons = 18 (from [Ar]) + 10 (3d) + 1 (4s) = 29 electrons.
2. Determining the Element:
The number of electrons in a neutral atom equals its atomic number. An element with 29 electrons has an atomic number of 29, which corresponds to copper (Cu) on the periodic table.
3. Verifying the Configuration:
Copper (Cu, atomic number 29) has the expected ground state configuration. The noble gas core [Ar] is followed by the filling of the 4s and 3d orbitals. Typically, for transition metals, the 4s orbital fills before 3d. However, copper is an exception:
- Expected: [Ar]4s\(^2\)3d\(^9\)
- Actual: [Ar]4s\(^1\)3d\(^{10}\)
This exception occurs because a fully filled 3d subshell (3d\(^{10}\)) is more stable than a partially filled one, so an electron from 4s moves to 3d, resulting in [Ar]3d\(^{10}\)4s\(^1\), which matches the given configuration.
Final Answer:
The element with the electronic configuration [Ar]3d\(^{10}\)4s\(^1\) is copper (Cu).
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.