The electronic configuration for sodium is 1s2 2s2 2p6 3s1, and the electronic configuration for potassium is 1s2 2s2 2p6 3s2 3p6 4s1. It requires 5.1 eV to remove the outermost electron (3s1) from sodium. What can you say about the energy required to remove the outermost electron from potassium (4s1)?
Since each is the outermost electron around the atom, the energy required will be equal to 5.1 eV. |
Since each electron is in an s orbital, the energy required will be equal to 5.1 eV. |
Since the outermost electron around sodium is in the 3s state, and the outermost electron around potassium is in the 4s state, the energy required will be greater than 5.1 eV. |
Since the outermost electron around sodium is in the 3s state, and the outermost electron around potassium is in the 4s state, the energy required will be less than 5.1 eV. |
none of the above |
Ans : The energy required to remove the outermost electron from potassium (4s1) will be less than 5.1 eV because of following reason :
Since the outermost electron around sodium is in the 3s state, and the outermost electron around potassium is in the 4s state, the energy required will be less than 5.1 eV. Hence the third option is correct one.
The electronic configuration for sodium is 1s2 2s2 2p6 3s1, and the electronic configuration for potassium...
The electronic configuration for boron is 1s2 2s2 2p1, and the electronic configuration for aluminum is 1s2 2s2 2p6 3s2 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be equal...
The electronic configuration for boron is 1s2 2s2 2p1, and the electronic configuration for aluminum is 1s2 2s2 2p6 3s2 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Group of answer choices Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the...
REFER TO THE ELECTRON CONFIGURATIONS SHOWN BELOW. (A)1s2 2s2 2p6 3s2 3p4 (B)1s2 2s2 2p6 3s2 3p6 4s1 3d5 (C)1s2 2s2 2p8 3s2 3p6 (D)1s2 2s2 2p6 3s2 3p6 4s2 3d4 (E)1s2 2s2 2p6 3s2 3p6 5 -electron configuration of Cr 6 -violates Pauli exclusion principle 7 -noble gas configuration with explaniation
Question 22 5 pts The electronic configuration for boron is 1s2 2s22p1, and the electronic configuration for aluminum is 152 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p 1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the...
The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s2 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 ev. Since each electron is in a p orbital, the energy required will be equal to 8.298...
The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be equal to...
A calcium atom (Ca) with Z-20 has the following electronic configurations: Ca: 1s2 2s2 2p6 3s2 3p6 4s2 Ca*: 1s2 2s2 2p6 3s2 3p6 4s1 3d Determine the term symbols arising from the above configurations indicating the values of L, S and J. (10 marks)
What is the ground state electron configuration of Rubidium ( Rb ) ? a) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 c) 5s1 b) 1s2 2s2 2p6 3s2 3p6 4s2 4p6 4d10 5s1 d) 1s2 2s2 2p6 3s2 3p6 3d10 4p6 5s2 5p2
Question 22 5 pts The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy...
Which of the following electronic configurations is not possible: Li-3: 1s2 2s1 N-7: 1s2 2s2 2p3 Ne-10: 1s2 2p6 3s2 K-19: 1s2 2s2 2p6 3s2 3p6 4s1 All of the above configurations are possible.