Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
| Orbitals | n | n + l |
| 4d | 4 | 6 |
| 4f | 4 | 7 |
| 5s | 5 | 5 |
| 5p | 5 | 6 |
The lowest energy orbital is 5s
APPEARS IN
संबंधित प्रश्न
Choose the correct option.
“No two electrons in the same atoms can have identical set of four quantum numbers”. This statement is known as -
Write electronic configurations of \[\ce{Fe, Fe2+, Fe3+}\].
Write condensed orbital notation of electronic configuration of the following element:
Chlorine (Z = 17)
Draw shapes of 2s orbitals.
Using the concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells, and orbitals.
Indicate the number of unpaired electron in:
Cr (Z = 24)
Which mineral among the following contains vanadium in it?
Which one of the following orders is CORRECT in case of energy of the given subshells?
P: n = 4; l = 3
Q: n = 5; I = 1
R: n = 5; l = 0
S: n = 4; l = 2
How many electrons in 19K have n = 3, l = 1?
Which of the following options does not represent ground state electronic configuration of an atom?
Total number of orbitals associated with third shell will be ______.
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 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 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 |
Choose the INCORRECT statement
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?
What are the possible values of magnetic quantum number for p orbital?
