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Indicate the number of unpaired electrons in Si (Z = 14). - Chemistry

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प्रश्न

Indicate the number of unpaired electrons in \[\ce{Si}\] (Z = 14).

लघु उत्तर

उत्तर

Electronic configuration of \[\ce{Si}\] (Z = 14) is 1s2 2s2 2p6 3s2 3p2

Orbital diagram:

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓  
1s 2s 2p 3s 3p

Number of unpaired electrons = 2

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पाठ 4: Structure of Atom - Exercises [पृष्ठ ५४]

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बालभारती Chemistry [English] 11 Standard
पाठ 4 Structure of Atom
Exercises | Q 6. (S)(a) | पृष्ठ ५४

संबंधित प्रश्‍न

Using s, p, d notations, describe the orbital with the following quantum numbers n = 1, l = 0.


Choose the correct option.

p-orbitals are _________ in shape.


Define the term Electronic configuration


State and explain Pauli’s exclusion principle.


Explain the anomalous behaviour of chromium.


Write orbital notations for the electron in orbitals with the following quantum numbers.

n = 4, l = 2


Write electronic configurations of \[\ce{Fe, Fe2+, Fe3+}\].


Write condensed orbital notation of electronic configuration of the following element:

Lithium (Z = 3)


Write condensed orbital notation of electronic configuration of the following element:

Silicon (Z = 14)


Explain in brief, the significance of the azimuthal quantum number.


If n = 3, what are the quantum number l and m?


How many electrons can fit in the orbital for which n = 4 and l = 2?


Which of the following has a greater number of electrons than neutrons?

(Mass number of Mg, C, O and Na is 24, 12, 16 and 23 respectively).


Which one of the following is NOT possible?


Which of the following options does not represent ground state electronic configuration of an atom?


The pair of ions having same electronic configuration is ______.


Nickel atom can lose two electrons to form \[\ce{Ni^{2+}}\] ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.


The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.

Based upon the above information, solve the questions given below:

Which of the following orbitals has the lowest energy?

4d, 4f, 5s, 5p


The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.

Based upon the above information, solve the questions given below:

Which of the following orbitals has the lowest energy?

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The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this configuration explained?


Match the following species with their corresponding ground state electronic configuration.

Atom / Ion Electronic configuration
(i) \[\ce{Cu}\] (a) 1s2 2s2 2p6 3s2 3p6 3d10
(ii) \[\ce{Cu^{2+}}\] (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2
(iii) \[\ce{Zn^{2+}}\] (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1
(iv) \[\ce{Cr^{3+}}\] (d) 1s2 2s2 2p6 3s2 3p6 3d9
  (e) 1s2 2s2 2p6 3s2 3p6 3d3

Match the following

(i) Photon (a) Value is 4 for N shell
(ii) Electron (b) Probability density
(iii) ψ2 (c) Always positive value
(iv) Principal quantum number n (d) Exhibits both momentum and wavelength

Match species given in Column I with the electronic configuration given in Column II.

Column I Column II
(i) \[\ce{Cr}\] (a) [Ar]3d84s0
(ii) \[\ce{Fe^{2+}}\] (b) [Ar]3d104s1
(iii) \[\ce{Ni^{2+}}\] (c) [Ar]3d64s0
(iv) \[\ce{Cu}\] (d) [Ar] 3d54s1
  (e) [Ar]3d64s2

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