Advertisements
Advertisements
प्रश्न
Write condensed orbital notation of electronic configuration of the following element:
Chlorine (Z = 17)
Advertisements
उत्तर
| Element | Condensed orbital notation |
| Chlorine (Z = 17) | [Ne] 3s2 3p5 |
APPEARS IN
संबंधित प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 3; l =1.
Choose the correct option.
“No two electrons in the same atoms can have identical set of four quantum numbers”. This statement is known as -
Choose the correct option.
Principal Quantum number describes -
Give the names of quantum numbers.
State the order of filling atomic orbitals following Aufbau principle.
Explain the anomalous behaviour of chromium.
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 2, l = 1
Write condensed orbital notation of electronic configuration of the following element:
Oxygen (Z = 8)
Write condensed orbital notation of electronic configuration of the following element:
Calcium (Z = 20)
Draw shapes of 2s orbitals.
Draw shapes of 2p orbitals.
Explain in brief, the significance of the azimuthal quantum number.
Write a note on ‘Principal Quantum number.
The principal quantum number (n) and magnetic quantum number (ml) for the valence electrons of rubidium atom (Z = 37) are ____________ respectively.
How many electrons can fit in the orbital for which n = 4 and l = 2?
The number of radial nodes for 3p orbital is ______.
Which of the following statements concerning the quantum numbers are correct?
(i) Angular quantum number determines the three dimensional shape of the orbital.
(ii) The principal quantum number determines the orientation and energy of the orbital.
(iii) Magnetic quantum number determines the size of the orbital.
(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.
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.
Calculate the total number of angular nodes and radial nodes present in 3p orbital.
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, arrange the following orbitals in the increasing order of energy.
4s, 3s, 3p, 4d
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 |
Which of the following is the correct plot for the probability density ψ2 (r) as a function of distance 'r' of the electron from the nucleus for 2s orbitals?
In assigning R - S configuration, which among the following groups has highest priority?
