The element 12Mg24 has 1s2 2s2 2p6 3s2 electronic configuration. The easiest method for Mg to reach an octet is by
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loosing the two electrons in 3s2 level to reach 2s2 2p6 valence electronic configuration |
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2. |
gaining 6 electrons to 3p level to reach 3p6configuration. With this acquisition the Mg will have a 3s2 and 3p6 valence shell configuration |
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3. |
Share 6 electrons with 6 F atoms |
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4. |
None of the above |
The easiest method for Mg to reach an octet is by loosing the two electrons in 3s2 level to reach 2s2 2p6 valence electronic configuration .it is easy for Mg to lose two outer electrons to reach an octet level.
The element 12Mg24 has 1s2 2s2 2p6 3s2 electronic configuration. The easiest method for Mg to...
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
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 is the complete electron configuration for what element with a negative one charge?
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)
1. Which element has the electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d5 ? a. Br b. Mn c. Nb d. Tc 2. What is the value of 27oC on the Fahrenheit scale? Group of answer choices a. -6.8 b. 106 c. 300 d. 81 e. none of these 3. Which of the following atoms has the largest radius? Group of answer choices a. S b. Se c. Br d. O
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...
From the consideration of electronic configuration, which of the elements indicated below would be classified as a transition element? 1s2 2s2 2p6 3s2 3p5 1s2 2s2 2p6 3s2 3p6 4s2 1s2 2s2 2p6 3s2 3p6 3d5 4s2 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
Q2: b) Three elements have the electron configurations: 1s' 2s2 2p 3s 3p 1s2 2s2 2p6 3s2 1s2 2s2 2p6 3s 3p 4s. The first ionization energies of these elements (not in the same order) are: 0.419, 0.735, 1.527 MI/mol (MJ for mega Jouls). The atomic radii are 1.60 A°, 0.98 A°, 2.35 A°. Identify the three elements and match the appropriate values of ionization energy and atomic radius to each configuration Element Atomic Radius lonization Energy 1. 2. 3.
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...
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7 An atom of chromium has unpaired electrons and is A) 3, diamagnetic 5, diamagnetic 3, paramagnetic 5, paramagnetic 6, paramagnetic B) C) D) E) 8 The electronic configuration and filling order of the element whose atomic number is 26 is: A) 1s2 2s2 2p6 3s2 3p6 4s0 3d8 B) 1s2 2s2 2p6 3s2 3p6 3d6 4s2 1s2 2s2 2p6 3s2 3p6...